AP Chemistry Unit 8: Acids and Bases — Free Review Games.

A Bronsted-Lowry acid is a proton (H+) donor. When it donates a proton to another substance, it acts as an acid.

pH = -log[H+]. If pH = 3, then [H+] = 10^-3 M = 0.001 M.

HNO3 is one of the six common strong acids that completely ionize in water. The others listed are weak acids.

A conjugate base is formed when an acid donates a proton. When H2O donates H+, it becomes OH-.

At 25 C, Kw = [H+][OH-] = 1.0e-14. In a neutral solution, [H+] = [OH-] = 1.0e-7 M, so pH = 7.

NaOH is a strong base: [OH-] = 0.010 M. pOH = -log(0.010) = 2. pH = 14 - 2 = 12.

\(K_a = x^2/0.10\) (assuming \(x \ll 0.10\)). \(x^2 = 1.8 \times 10^{-6}\). \(x = [H^+] = 1.34 \times 10^{-3}\). \(pH = -\log(1.34 \times 10^{-3}) = 2.87\).

A buffer is most effective when pH = pKa, which occurs when [HA] = [A-]. At this point, the buffer has maximum capacity to neutralize added acid or base.

At the equivalence point, all weak acid has been converted to its conjugate base (A-), which hydrolyzes to produce OH-. The solution is basic (pH > 7).

pH = pKa + log([A-]/[HA]) relates pH to the pKa and ratio of conjugate base to acid.

pH = pKa + log([A-]/[HA]) = 4.74 + log(0.30/0.20) = 4.74 + log(1.5) = 4.74 + 0.18 = 4.92.

At the half-equivalence point, [HA] = [A-]. By Henderson-Hasselbalch, pH = pKa + log(1) = pKa = 4.74.

Phenolphthalein changes color at pH 8.2-10.0, which brackets the equivalence point pH of 8.7.

CO3(2-) is the conjugate base of weak acid HCO3-. It reacts with water: CO3(2-) + H2O <-> HCO3- + OH-, producing hydroxide and making the solution basic.

Ka * Kb = Kw = 1.0e-14. Kb = Kw/Ka = 1.0e-14 / 6.8e-4 = 1.47e-11, approximately 1.5e-11.

The Arrhenius definition states that a base produces OH- ions in aqueous solution (e.g., NaOH → Na+ + OH-). Choice A describes the Bronsted-Lowry definition of a base, and choice C describes a Lewis base — both are broader definitions developed later to cover non-aqueous contexts.

HF (hydrofluoric acid) is a weak acid with Ka = 6.8 x 10^-4; it only partially ionizes in water. HCl, HNO3, and HI are all strong acids that dissociate completely. Despite fluorine being the most electronegative element, the H-F bond is unusually strong, which limits ionization.

At 25°C, pH + pOH = 14. Therefore pH = 14 - 5 = 9. This solution is basic because pH > 7. Choice A would mean pH equals pOH, which only occurs at neutrality (pH = pOH = 7). Choice D is the sum, not the difference.

A conjugate acid is formed when a base gains a proton (H+). NH3 gains one H+ to become NH4+ (ammonium). NH2- is the conjugate base of NH3 (formed by losing a proton). N2H4 (hydrazine) is a structurally different compound, not a conjugate.

KOH (potassium hydroxide) is a strong base that dissociates completely in water. NH3 and CH3NH2 are weak bases that only partially accept protons. Al(OH)3 is amphoteric and only sparingly soluble — it does not fully dissociate and is not classified as a strong base.

Kw = [H+][OH-] = 1.0 x 10^-14 at 25°C. This is the equilibrium expression for the autoionization of water: H2O ⇌ H+ + OH-. [H2O] is omitted because it is a pure liquid. In a neutral solution, [H+] = [OH-] = 1.0 x 10^-7 M, but this is the individual concentration, not the expression for Kw.

NaOH is a strong base that fully dissociates to produce OH- ions. Adding OH- increases [OH-], decreases [H+] (since Kw = [H+][OH-] is constant), and raises the pH above 7. Na+ is a spectator ion with no effect on pH. Water has no buffering capacity on its own.

HCl is a strong acid that fully dissociates, so [H+] = 0.050 M. pH = -log(0.050) = -log(5.0 x 10^-2) = -(log 5.0 + log 10^-2) = -(0.699 - 2) = 1.30. Choice A (pH 1.00) would correspond to [H+] = 0.10 M. Choice D is the pOH, not the pH.

[H+] = 0.013 x 0.10 = 1.3 x 10^-3 M. Using the equilibrium expression and the approximation that [HA]eq ≈ 0.10 M: Ka = [H+][A-] / [HA] = (1.3 x 10^-3)^2 / 0.10 = 1.69 x 10^-6 / 0.10 = 1.7 x 10^-5. Choice A is just [H+], not Ka. Choice D is off by a factor of 10.

pOH = -log[OH-] = -log(2.5 x 10^-3). log(2.5) ≈ 0.40, so pOH = -(0.40 - 3) = 2.60. This gives pH = 14 - 2.60 = 11.40 (choice C is the pH, not pOH — a common error). Choice B of 3.40 would correspond to [OH-] = 4 x 10^-4 M.

A buffer requires a weak acid and its conjugate base (or weak base and conjugate acid) in significant amounts. CH3COOH (acetic acid) is a weak acid and CH3COONa provides its conjugate base CH3COO-, making an effective buffer. HCl is a strong acid whose conjugate base Cl- is negligible. Equal moles of HCl and NaOH neutralize each other, yielding a neutral salt solution with no buffering capacity.

Adding CH3COO- (the conjugate base) shifts the equilibrium CH3COOH ⇌ H+ + CH3COO- to the left by Le Chatelier's principle. This is the common ion effect: the common ion CH3COO- suppresses dissociation of acetic acid, reducing [H+] and lowering percent ionization. Ka is constant but the ratio of products to reactants changes.

At the half-equivalence point, exactly half the weak acid has been neutralized, so [HA] = [A-]. Substituting into the Henderson-Hasselbalch equation: pH = pKa + log([A-]/[HA]) = pKa + log(1) = pKa. Choice B is wrong because pH = pKa, not Ka (different units entirely). This point is used experimentally to determine pKa.

For any conjugate acid-base pair, Ka x Kb = Kw = 1.0 x 10^-14 at 25°C. This follows because the two equilibria (HA ⇌ H+ + A- and A- + H2O ⇌ HA + OH-) sum to the autoionization of water. This relationship allows calculation of the Kb of a conjugate base if Ka of the acid is known, and vice versa.

A Lewis acid is an electron pair acceptor. BF3 has an incomplete octet (only 6 electrons around boron) and accepts the electron pair from F- to form BF4-. F- is the Lewis base (electron pair donor). BF4- is the product. This reaction illustrates that Lewis acid-base theory extends beyond proton transfer to include electron pair interactions.

Using the approximation: \([OH^-] = \sqrt{K_b \times C} = \sqrt{1.8 \times 10^{-5} \times 0.10} = \sqrt{1.8 \times 10^{-6}} = 1.34 \times 10^{-3}\) M. $pOH = -\log(1.34 \times 10^{-3}) = 2.87$. \(pH = 14 - 2.87 = 11.13\). Choice B (\(8.87\)) is the pH of a much more dilute or weaker base solution. Choice A is the pOH, not the pH — a common sign error.

Because Ka1 (1.0 x 10^-3) is 100,000 times larger than Ka2 (1.0 x 10^-8), the first ionization produces far more H+ than the second. In practice, [H+] from the second step is negligible. Ka2 is always smaller than Ka1 for polyprotic acids because removing a second proton from an already-negative ion requires overcoming additional electrostatic repulsion.

NH4Cl dissociates to give NH4+ and Cl-. NH4+ is the conjugate acid of the weak base NH3, so it undergoes hydrolysis: NH4+ + H2O ⇌ NH3 + H3O+, producing an acidic solution. KCl and NaCl are neutral salts (strong acid + strong base). NaCH3COO and Na2CO3 produce basic solutions because their anions are conjugate bases of weak acids.

Using Henderson-Hasselbalch: pH = pKa + log([base]/[acid]) = 9.26 + log(0.30/0.20) = 9.26 + log(1.50) = 9.26 + 0.18 = 9.44. The base here is NH3 and the acid is NH4+. Choice B (9.08) uses the ratio inverted. Choice A (8.92) is 9.26 - 0.18 x 2, a common arithmetic error. When [base] > [acid], the pH must be above pKa, confirming choice C.

pKa = -log(1.8 x 10^-5) = 4.74. Adding 0.050 mol HCl converts 0.050 mol CH3COO- to CH3COOH: new [CH3COOH] = 0.30 + 0.050 = 0.35 M; new [CH3COO-] = 0.50 - 0.050 = 0.45 M. pH = 4.74 + log(0.45/0.35) = 4.74 + log(1.286) = 4.74 + 0.11 = 4.85. The pH dropped slightly from the original (4.74 + log(0.50/0.30) = 4.96) but remains well-buffered, demonstrating buffer action.

At the equivalence point, all \(NH_3\) is converted to \(NH_4^+\). Moles \(NH_3 = 0.0250\) L \(\times 0.200\) M \(= 0.00500\) mol. Volume at equivalence \(= 50.0\) mL, so \([NH_4^+] = 0.00500/0.0500 = 0.100\) M. \(K_a(NH_4^+) = K_w/K_b = 1.0 \times 10^{-14} / 1.8 \times 10^{-5} = 5.6 \times 10^{-10}\). \([H^+] = \sqrt{5.6 \times 10^{-10} \times 0.100} = 7.5 \times 10^{-6}\) M. \(pH = -\log(7.5 \times 10^{-6}) \approx 5.13\). The solution is acidic because \(NH_4^+\) is a weak acid — the equivalence point of a weak base/strong acid titration is always below pH 7.

Each additional electronegative oxygen atom on the central chlorine atom withdraws electron density inductively along the chain toward the O-H bond. This polarizes the O-H bond more strongly, lowering electron density on oxygen and making it easier to release the proton. This principle — more oxygens on the central atom increases oxoacid strength — applies broadly (e.g., HClO < HClO2 < HClO3 < HClO4). Chlorine's electronegativity does not change between compounds.

A neutral solution requires \([H^+] = [OH^-]\). From \(K_w = [H^+][OH^-] = 2.4 \times 10^{-14}\), we get \([H^+] = \sqrt{2.4 \times 10^{-14}} = 1.549 \times 10^{-7}\) M. \(pH = -\log(1.549 \times 10^{-7}) = 6.81\). Neutrality is defined as \([H^+] = [OH^-]\), not as \(pH = 7\). Since \(K_w\) increases with temperature (autoionization is endothermic), the neutral pH is lower than \(7.00\) at body temperature. This does not mean blood is acidic — it is still neutral because \([H^+] = [OH^-]\).

Moles $HCOOH = 0.100$ L \(\times 0.050\) M \(= 0.0050\) mol. Volume NaOH \(= 0.0050\) mol \(/ 0.10\) M \(= 0.050\) L \(= 50.0\) mL. At equivalence, all formic acid becomes formate ($HCOO^-$). Total volume \(= 150.0\) mL; $[HCOO^-] = 0.0050/0.150 = 0.0333$ M. $K_b(HCOO^-) = K_w/K_a = 1.0 \times 10^{-14} / 1.8 \times 10^{-4} = 5.6 \times 10^{-11}$. \([OH^-] = \sqrt{5.6 \times 10^{-11} \times 0.0333} = 1.36 \times 10^{-6}\) M. $pOH = 5.87$; \(pH = 14 - 5.87 = 8.13\). The equivalence point is basic because formate is the conjugate base of a weak acid.

\(NH_4Cl\) is a salt of a weak base (\(NH_3\)) and strong acid (HCl). In solution, \(NH_4^+\) hydrolyzes: \(NH_4^+ \rightleftharpoons NH_3 + H^+\). \(K_a(NH_4^+) = K_w / K_b(NH_3) = 1.0 \times 10^{-14} / 1.8 \times 10^{-5} = 5.6 \times 10^{-10}\). \([H^+] = \sqrt{K_a \times C} = \sqrt{5.6 \times 10^{-10} \times 0.10} = \sqrt{5.6 \times 10^{-11}} = 7.5 \times 10^{-6}\) M. \(pH = -\log(7.5 \times 10^{-6}) \approx 5.13\). Choice D (\(9.26\)) is the \(pK_a\) of \(NH_4^+\), not the pH of the solution.

A Bronsted-Lowry acid is defined as a proton donor — it transfers H+ to a base during a reaction. Choice B describes a Bronsted-Lowry base. Choices C and D describe Lewis acids and bases, which are defined in terms of electron pairs rather than proton transfer. The Bronsted-Lowry model is broader than Arrhenius (which requires OH- or H+ directly) but narrower than Lewis.

HF is a weak acid with Ka ≈ 6.8 × 10^-4; it does not fully dissociate in aqueous solution. HCl, HNO3, and HBr are all strong acids that dissociate essentially 100% in water. Despite fluorine being highly electronegative, the H-F bond has a very high bond dissociation energy that limits ionization, making HF the only weak acid in this list.

HCl is a strong acid that dissociates completely, so [H+] = 0.010 M = 1.0 × 10^-2 M. pH = -log(1.0 × 10^-2) = 2. Choice A (pH = 1) corresponds to a 0.10 M HCl solution, which is 10 times more concentrated. Choice D (pH = 12) describes a basic solution, which HCl cannot produce.

A conjugate base forms when an acid donates exactly one proton. H2PO4- loses one H+ to form HPO4^2-. H3PO4 is actually the conjugate acid of H2PO4- (it has one more proton). PO4^3- results from losing two protons from H2PO4-, which skips a step and is not a conjugate base relationship. The conjugate relationship always involves a one-proton difference.

An amphoteric species can act as either an acid or a base depending on its environment. HCO3- (bicarbonate) can donate a proton to act as an acid (HCO3- → H+ + CO3^2-) or accept a proton to act as a base (HCO3- + H+ → H2CO3). HCl acts only as an acid, NaOH acts only as a base, and NaCl is a neutral salt that does not participate in proton transfer equilibria.

At \(25°C\), \(K_w = [H^+][OH^-] = 1.0 \times 10^{-14}\). In pure water, autoionization produces equal amounts of \(H^+\) and \(OH^-\), so \([OH^-] = \sqrt{1.0 \times 10^{-14}} = 1.0 \times 10^{-7}\) M. Choice A (\(1.0 \times 10^{-14}\)) is the value of \(K_w\) itself, not the concentration of \(OH^-\). Values of \(7.0 \times 10^{-1}\) M and \(1.0\) M would represent implausibly high concentrations of hydroxide in neutral water.

A larger Ka value indicates a greater extent of dissociation in water and therefore a stronger acid. Ka = 1.0 × 10^-2 is the largest value listed, making that acid the strongest. Acid strength increases as Ka increases (or as pKa decreases). Ka = 1.0 × 10^-10 describes the weakest acid in the list, dissociating to the smallest extent.

At 25°C, the relationship pH + pOH = 14.00 always holds, because pKw = 14. If pH = 3.0, then pOH = 14.0 - 3.0 = 11.0. Choice A (pOH = 3.0) would imply the solution is simultaneously strongly acidic and strongly basic with equal [H+] and [OH-], which only applies to neutral solutions. Choice D (pOH = 14) corresponds to an extremely basic solution with pH near zero.

For a weak acid, \([H^+] = \sqrt{K_a \times C} = \sqrt{1.8 \times 10^{-5} \times 0.10} = \sqrt{1.8 \times 10^{-6}} \approx 1.34 \times 10^{-3}\) M. \(pH = -\log(1.34 \times 10^{-3}) \approx 2.87\), approximately \(2.9\). Choice C (\(4.7\)) is the \(pK_a\) of acetic acid — this equals the pH only at the half-equivalence point of a titration when $[CH_3COOH] = [CH_3COO^-]$, not for a pure weak acid solution. Choice A (\(1.0\)) would correspond to a \(0.10\) M strong acid solution.

Diluting a weak acid increases its percent dissociation. As concentration decreases, the equilibrium shifts right (Le Chatelier), but [H+] falls more slowly than [HA], so the fraction dissociated increases. Increasing concentration (Choice A) actually decreases percent dissociation. Adding the conjugate base salt (Choice B) suppresses dissociation via the common-ion effect. Decreasing temperature for an exothermic dissociation (Choice D) shifts the equilibrium left, reducing dissociation.

The Henderson-Hasselbalch equation is pH = pKa + log([A-]/[HA]). Setting pH = pKa + 1: pKa + 1 = pKa + log([A-]/[HA]), so log([A-]/[HA]) = 1, giving [A-]/[HA] = 10^1 = 10. A ratio of 0.1 gives pH = pKa - 1. A ratio of 1 gives pH = pKa. A ratio of 100 gives pH = pKa + 2. These relationships show that buffers work best within one pH unit of the pKa.

At the equivalence point of a weak base/strong acid titration, all the weak base has been converted to its conjugate acid. This conjugate acid is a weak acid that partially donates protons to water, producing a slightly acidic solution with pH below 7. The solution would be exactly pH 7 only if both the acid and base were strong. Choice C (pH greater than 7) describes the equivalence point of a weak acid/strong base titration, not a weak base/strong acid titration.

The equilibrium constant K = Ka(CH3COOH) / Ka(HCN) = (1.8 × 10^-5) / (6.2 × 10^-10) ≈ 2.9 × 10^4, which is much greater than 1. Acid-base equilibria always favor the production of the weaker acid and weaker base. Since acetic acid (Ka = 1.8 × 10^-5) is much stronger than HCN (Ka = 6.2 × 10^-10), the reaction proceeds forward to form the weaker acid HCN. Choice D has the reasoning reversed — CN- is actually a stronger base than CH3COO- because it is the conjugate base of the weaker acid.

When \(K_w\) increases with temperature, \([H^+] = [OH^-] = \sqrt{K_w}\) both increase, so \(pH = -\log[H^+]\) drops below \(7\). However, since \([H^+]\) still equals \([OH^-]\), the solution remains chemically neutral — neutral means equal concentrations of \(H^+\) and \(OH^-\), not necessarily \(pH = 7\). \(pH = 7\) is only the neutrality point at \(25°C\). Choice A incorrectly equates 'pH < 7' with 'acidic'; Choice C incorrectly assumes neutrality always means \(pH = 7\).

H2SO4 first ionization is complete (Ka1 >> 1), yielding [H+] = 0.10 M and [HSO4-] = 0.10 M initially. For the second ionization, Ka2 = 0.012 is significant but not >> 1, so HSO4- only partially dissociates, contributing additional H+ and producing SO4^2-. The result is [H+] somewhere between 0.10 and 0.20 M, with both HSO4- and SO4^2- present. Choice B (complete double ionization) overestimates by assuming Ka2 is much greater than 1.

The Henderson-Hasselbalch equation relies on the approximation that equilibrium concentrations of acid and conjugate base are close to their initial values. This assumption is most valid when [A-]/[HA] is between 0.1 and 10 (pH within ±1 unit of pKa), because the extent of further reaction is small relative to the buffering species present. Very low concentrations (Choice B) worsen the approximation because the autoionization of water becomes non-negligible. The equation requires a weak acid (not strong, Choice C), and Ka equaling Kw (Choice D) has no special significance for buffer accuracy.

When hydroxide ions are added to an acetate buffer, the weak acid component (CH3COOH) reacts with the OH- to form acetate and water: CH3COOH + OH- → CH3COO- + H2O. This consumes the base before it can significantly raise [OH-] in solution. Choice B is chemically incorrect — acetate (a weak base) cannot displace OH- to form oxide ions. Choice C is incorrect because Na+ is a spectator ion and NaOH does not precipitate. The buffer works precisely because the weak acid provides a reservoir to neutralize added base.

At the equivalence point, \(50.0\) mL of \(0.100\) M NaOH is added, giving total volume \(= 100.0\) mL. All acetic acid is converted to acetate: $[CH_3COO^-] = (0.0500 \text{ L} \times 0.100 \text{ mol/L}) / 0.100 \text{ L} = 0.0500$ M. $K_b(CH_3COO^-) = K_w/K_a = (1.0 \times 10^{-14}) / (1.8 \times 10^{-5}) = 5.6 \times 10^{-10}$. \([OH^-] = \sqrt{5.6 \times 10^{-10} \times 0.0500} = \sqrt{2.8 \times 10^{-11}} \approx 5.3 \times 10^{-6}\) M. $pOH \approx 5.28$, so \(pH = 14.00 - 5.28 = 8.72\). The pH exceeds \(7\) because acetate is the conjugate base of a weak acid and hydrolyzes to produce a basic solution.

pH = 3.00 means [H+] = 1.0 × 10^-3 M. From the equilibrium HA ⇌ H+ + A-, [H+] = [A-] = 1.0 × 10^-3 M at equilibrium, and [HA] = 0.100 - 0.001 = 0.099 M. Ka = [H+][A-] / [HA] = (1.0 × 10^-3)^2 / 0.099 = 1.0 × 10^-6 / 0.099 ≈ 1.01 × 10^-5. Choice A (1.0 × 10^-3) is simply [H+] — a common error is confusing Ka with [H+]. Ka must incorporate the undissociated acid concentration in the denominator.

The added H+ reacts completely with acetate: CH3COO- + H+ → CH3COOH. After reaction: moles CH3COO- = 0.20 - 0.050 = 0.150 mol; moles CH3COOH = 0.20 + 0.050 = 0.250 mol. Using Henderson-Hasselbalch: pH = 4.74 + log(0.150 / 0.250) = 4.74 + log(0.600) = 4.74 - 0.222 = 4.52. Choice B (4.74) was the initial buffer pH before acid was added. Adding acid consumes conjugate base and forms more weak acid, which shifts the ratio and lowers the pH.

A naive calculation gives \(pH = -\log(1.0 \times 10^{-8}) = 8\), but this is wrong — an acid cannot make a solution basic. At such a low HCl concentration, the \(H^+\) from water autoionization (\(\sim 10^{-7}\) M) is not negligible. Setting total \([H^+] = x\), charge balance gives: \(x^2 - (1.0 \times 10^{-8})x - 1.0 \times 10^{-14} = 0\). Solving: \(x \approx 1.05 \times 10^{-7}\) M, so \(pH = -\log(1.05 \times 10^{-7}) \approx 6.98\). The solution is very slightly acidic (just below pH \(7\)), not basic. Choice C (\(7.00\)) ignores the real though small contribution from HCl.

$NaHA$ dissolves to give the amphiprotic intermediate \(HA^-\), which can both donate and accept protons. For an amphiprotic species at reasonable concentration, $[H^+] \approx \sqrt{K_{a1} \times K_{a2}} = \sqrt{1.0 \times 10^{-4} \times 1.0 \times 10^{-9}} = \sqrt{1.0 \times 10^{-13}} = 10^{-6.5} \approx 3.2 \times 10^{-7}$ M, giving \(pH \approx 6.5\). Equivalently, \(pH \approx (pK_{a1} + pK_{a2}) / 2 = (4 + 9) / 2 = 6.5\). This result is relatively independent of concentration. Choice B (\(4.5\)) incorrectly uses only \(K_{a1}\), ignoring the base behavior of \(HA^-\).

When pH = pKa = 5.00, the Henderson-Hasselbalch equation requires log([A-]/[HA]) = 0, so [A-] = [HA]. This is the half-equivalence point, where exactly half the weak acid has been converted to conjugate base. Total moles HA = 0.0250 L × 0.100 mol/L = 2.50 × 10^-3 mol. At the half-equivalence point, 1.25 × 10^-3 mol NaOH is needed. Volume = 1.25 × 10^-3 mol / 0.100 mol/L = 0.0125 L = 12.5 mL. Choice C (25.0 mL) reaches the full equivalence point, where pH is greater than 7, not equal to pKa.

Using the approximation \([H^+] = \sqrt{K_a \times C}\): \([H^+] = \sqrt{1.0 \times 10^{-6} \times 0.020} = \sqrt{2.0 \times 10^{-8}} \approx 1.41 \times 10^{-4}\) M. Percent dissociation $= ([H^+] / [HA]_{initial}) \times 100 = (1.41 \times 10^{-4} / 0.020) \times 100 = 0.71\%$. The approximation is valid because \(0.71\%\) is much less than \(5\%\). Choice C (\(7.1\%\)) results from a factor-of-10 arithmetic error. Choice A (\(0.071\%\)) results from using \(K_a/C\) instead of \(\sqrt{K_a/C}\) in the percent formula.

Applying Henderson-Hasselbalch with NH4+ as the acid and NH3 as its conjugate base: pH = pKa + log([NH3]/[NH4+]). Substituting: 9.00 = 9.26 + log([NH3]/[NH4+]). log([NH3]/[NH4+]) = -0.26. [NH3]/[NH4+] = 10^-0.26 ≈ 0.55. Because the target pH (9.00) is below the pKa (9.26), more conjugate acid (NH4+) than base (NH3) is needed, so the ratio is less than 1. Choice B (1.00) gives pH = pKa = 9.26. Choice C (1.82) is the reciprocal of the correct answer and would be used if the question asked for [NH4+]/[NH3].

The Bronsted-Lowry definition defines an acid as a proton (H⁺) donor and a base as a proton acceptor. This is broader than the Arrhenius definition, which only applies to aqueous solutions. Choice B describes a Bronsted-Lowry base, not an acid.

Hydrobromic acid (HBr) is one of the seven common strong acids that dissociate essentially 100% in dilute aqueous solution. Acetic acid, hydrofluoric acid, and phosphoric acid are all weak acids that only partially dissociate, establishing equilibria in solution.

HCl is a strong acid that dissociates completely, so [H⁺] = 0.010 M = 1.0 × 10⁻² M. pH = -log(1.0 × 10⁻²) = 2. Choice D (12) is the pOH value, not the pH — a common error from confusing pH and pOH.

A conjugate acid-base pair consists of two species that differ by exactly one proton. NH₄⁺ and NH₃ differ by one H⁺ (NH₄⁺ is the conjugate acid of NH₃). H₂SO₄ and SO₄²⁻ differ by two protons, so they are not a conjugate pair. HCl and NaCl share no proton-transfer relationship.

For any conjugate acid-base pair, Ka × Kb = Kw = 1.0 × 10⁻¹⁴ at 25°C. This relationship lets you calculate the Kb of a conjugate base from the Ka of the acid, or vice versa. The product does not equal 1; that would only hold if Ka = Kb, which is a special case.

Sodium acetate dissociates to give the acetate ion (CH₃COO⁻), the conjugate base of a weak acid. Acetate undergoes hydrolysis: CH₃COO⁻ + H₂O ⇌ CH₃COOH + OH⁻, producing excess OH⁻ and a basic solution. HCl is acidic, NaCl is neutral (both ions are conjugates of strong species), and acetic acid is acidic.

HCO₃⁻ (bicarbonate) is amphiprotic: it can donate a proton to act as an acid (HCO₃⁻ → CO₃²⁻ + H⁺) or accept a proton to act as a base (HCO₃⁻ + H⁺ → H₂CO₃). Cl⁻ and SO₄²⁻ are conjugate bases of strong acids and have negligible proton-accepting ability. Na⁺ is a spectator ion with no acid-base reactivity.

At pH 3.00, [H⁺] = 1.0 × 10⁻³ M. Each HA that dissociates produces one H⁺, so the dissociated concentration is also 1.0 × 10⁻³ M. Percent dissociation = (1.0 × 10⁻³ / 0.10) × 100% = 1.0%. Choice D (10%) results from incorrectly treating pH as percent, and choice C confuses the pH value with the percent.

Methylamine is a weak base. Using the approximation Kb ≈ x²/C: x = √(4.4 × 10⁻⁴ × 0.10) = 6.6 × 10⁻³ M = [OH⁻]. pOH = -log(6.6 × 10⁻³) ≈ 2.18, so pH = 14.00 - 2.18 ≈ 11.8. Choice A (2.2) is the pOH itself — a common error of stopping at pOH without converting to pH.

This is the common ion effect. Adding acetate ion (CH₃COO⁻) to the acetic acid equilibrium CH₃COOH ⇌ H⁺ + CH₃COO⁻ shifts the reaction to the left (Le Chatelier's principle), decreasing dissociation and lowering [H⁺]. Choice C is incorrect because Ka is constant but the equilibrium position shifts — Ka governs the ratio of concentrations, not their absolute values.

At the half-equivalence point, exactly half the acid has been converted to its conjugate base, so [HA] = [A⁻]. Substituting into Henderson-Hasselbalch: pH = pKa + log([A⁻]/[HA]) = pKa + log(1) = pKa. This is how weak acid pKa values are determined experimentally from titration curves. The pH equals 7.00 only if pKa happens to be 7, which is rarely the case.

At pH 4.00, [H⁺] = [A⁻] = 1.0 × 10⁻⁴ M. The equilibrium concentration of HA is 0.050 - 1.0 × 10⁻⁴ ≈ 0.0499 M. Ka = [H⁺][A⁻]/[HA] = (1.0 × 10⁻⁴)² / 0.0499 ≈ 2.0 × 10⁻⁷. Choice A (2.0 × 10⁻⁴) results from the error Ka = [H⁺]/[HA], omitting one factor of [H⁺] from the numerator.

A buffer requires a weak acid and its conjugate base (or a weak base and its conjugate acid) both present at significant concentrations. NH₃ (weak base) and NH₄Cl (which provides NH₄⁺, the conjugate acid) form a proper buffer pair. HCl is a strong acid with no weak conjugate partner in NaCl. Mixing NaOH with NH₃ gives only a weak base with no conjugate acid present.

pOH = -log(3.16 × 10⁻⁵) = 4.50. At 25°C, pH + pOH = 14.00, so pH = 14.00 - 4.50 = 9.50. The basic pH is expected because F⁻ is the conjugate base of the weak acid HF and partially hydrolyzes water to produce OH⁻. Choice A (4.50) is the pOH value — stopping at pOH without converting to pH is a classic error.

The reaction NH₃ + HCl → NH₄Cl goes to completion, consuming both reactants entirely. The resulting NH₄⁺ is the conjugate acid of the weak base NH₃ and hydrolyzes water: NH₄⁺ + H₂O ⇌ NH₃ + H₃O⁺, producing an acidic solution. The solution is not neutral because NH₄⁺ is a weak acid. Choice D is incorrect because all HCl has been consumed — none remains in solution.

Acid strength increases with larger Ka. HA dissociates most, giving the highest [H⁺] and lowest pH. HC dissociates the least, giving the lowest [H⁺] and therefore the highest pH. Equal initial concentration does not mean equal pH; the equilibrium degree of dissociation (governed by Ka) determines [H⁺]. Choice D confuses initial concentration with equilibrium concentration of H⁺.

Diluting a weak acid lowers the overall [H⁺], so pH increases. At the same time, dilution shifts the equilibrium HA ⇌ H⁺ + A⁻ to the right by Le Chatelier's principle, so a larger fraction of the acid dissociates — percent dissociation increases. Choice C is incorrect because although Ka is constant, the equilibrium position does shift upon dilution, which changes the percent dissociation.

An indicator should change color within the steep region of the titration curve surrounding the equivalence point. Since the equivalence point is at pH 9, phenolphthalein (transition range pH 8.2–10.0) brackets that point and produces a sharp color change at the correct moment. Methyl orange, bromocresol green, and methyl red all transition at pH 3–6, far below the equivalence point, and would give a premature, inaccurate endpoint.

Acetate ion hydrolyzes water: CH₃COO⁻ + H₂O ⇌ CH₃COOH + OH⁻. Kb = Kw/Ka = (1.0 × 10⁻¹⁴)/(1.8 × 10⁻⁵) = 5.6 × 10⁻¹⁰. Using [OH⁻] = √(Kb × C) = √(5.6 × 10⁻¹¹) = 7.5 × 10⁻⁶ M. pOH = -log(7.5 × 10⁻⁶) = 5.13, so pH = 14.00 - 5.13 ≈ 8.9. Choice A (5.1) is the pOH, not the pH — stopping at pOH is a common sign-inversion error.

Moles of acetic acid = 0.0500 L × 0.200 mol/L = 0.0100 mol. Moles of NaOH = 0.0250 L × 0.200 mol/L = 0.00500 mol. NaOH neutralizes half the acid: 0.00500 mol CH₃COOH remains and 0.00500 mol CH₃COO⁻ is produced. Because [HA] = [A⁻], the ratio [A⁻]/[HA] = 1, and by Henderson-Hasselbalch: pH = pKa + log(1) = 4.74. This is the half-equivalence point. Choice C (7.00) incorrectly assumes any acid-base mixing produces a neutral solution.

Added H⁺ reacts with the conjugate base: H⁺ + CH₃COO⁻ → CH₃COOH. Updated moles: CH₃COOH = 0.300 + 0.030 = 0.330 mol; CH₃COO⁻ = 0.300 - 0.030 = 0.270 mol. pH = 4.74 + log(0.270/0.330) = 4.74 + log(0.818) = 4.74 - 0.087 ≈ 4.65. Choice B (4.74) is the original buffer pH — it is incorrect because it ignores the shift in the [A⁻]/[HA] ratio caused by adding HCl.

HCO₃⁻ is an amphiprotic intermediate in a polyprotic system. Its pH is approximated by pH ≈ (pKa1 + pKa2)/2. pKa1 = -log(4.3 × 10⁻⁷) = 6.37 and pKa2 = -log(4.7 × 10⁻¹¹) = 10.33. pH ≈ (6.37 + 10.33)/2 = 8.35. Choice A (6.37) is pKa1, which corresponds to the half-equivalence point of the first dissociation step, not the amphiprotic intermediate.

Buffer capacity is the quantity of strong acid or base a buffer can absorb before a significant pH change occurs. Although both buffers share the same pH and the same [A⁻]/[HA] ratio, Buffer Y contains 100 times more moles of each component per liter. It can neutralize far more added H⁺ or OH⁻ before its ratio shifts enough to change pH significantly. The ratio governs pH; the absolute molar amounts govern capacity.

The simplifying approximation is valid only when x/C < 5%, requiring Ka/C ≪ 1. Here Ka/C = 1.0, so the approximation fails entirely — the result implies 100% dissociation, which is impossible for a weak acid. Solving the full quadratic x² + (1.0 × 10⁻³)x - (1.0 × 10⁻⁶) = 0 gives x = 6.18 × 10⁻⁴ M, so pH = -log(6.18 × 10⁻⁴) ≈ 3.21. The correct [H⁺] is substantially lower than the approximation predicted.

Moles NH₃ = 0.0300 L × 0.100 mol/L = 0.00300 mol. Moles HCl added = 0.0200 L × 0.100 mol/L = 0.00200 mol. After reaction: NH₄⁺ = 0.00200 mol, NH₃ = 0.00100 mol remain. For the NH₄⁺/NH₃ buffer: pKa(NH₄⁺) = -log(Kw/Kb) = -log(5.6 × 10⁻¹⁰) = 9.26. pH = 9.26 + log([NH₃]/[NH₄⁺]) = 9.26 + log(0.00100/0.00200) = 9.26 - 0.30 = 8.96. Choice A (9.26) is the pKa of NH₄⁺, which would be the pH only at the half-equivalence point where moles NH₃ equal moles NH₄⁺.

HClO₄ (perchloric acid) is one of the seven strong acids that dissociate completely in water. HF is a weak acid despite containing a halogen — its H–F bond is strong enough to prevent complete ionization, giving Ka ≈ 6.8 × 10⁻⁴. CH₃COOH and HCN are both weak acids with small Ka values.

HCl is a strong acid that dissociates completely, so [H⁺] = 0.010 M = 1.0 × 10⁻² M. pH = −log(1.0 × 10⁻²) = 2. A pH of 1 would correspond to 0.10 M HCl. Choices 12 and 13 would indicate basic solutions, which is impossible for an HCl solution.

A conjugate acid-base pair must differ by exactly one proton. H₂SO₄ and SO₄²⁻ differ by two protons, so they are not a conjugate pair. The true conjugate base of H₂SO₄ is HSO₄⁻. HCl/Cl⁻, NH₄⁺/NH₃, and H₂O/OH⁻ each differ by exactly one proton and are valid conjugate acid-base pairs.

Ka = 10^(−pKa) = 10^(−4.75). This equals 10^(−5) × 10^(0.25) = 1.0 × 10⁻⁵ × 1.78 = 1.78 × 10⁻⁵. The distractor 4.75 × 10⁻⁵ represents a common error where students use the pKa digits directly without computing the antilogarithm. The distractor 1.78 × 10⁻⁴ is off by one order of magnitude.

At 25°C, pH + pOH = 14.00. Therefore pOH = 14.0 − 9.0 = 5.0. A pOH of 9.0 would imply pH = 5.0, indicating an acidic solution, which contradicts the given information. A pOH of 7.0 would mean pH = 7.0, a neutral solution.

The Arrhenius definition defines a base as a substance that produces hydroxide ions (OH⁻) when dissolved in water, for example NaOH → Na⁺ + OH⁻. Choices A and B describe the Brønsted-Lowry model, which is broader and applies to non-aqueous systems as well. Choice D describes an Arrhenius acid.

Moles of acetic acid = 0.100 L × 0.10 M = 0.010 mol. Moles of NaOH = 0.050 L × 0.10 M = 0.005 mol. The NaOH converts exactly half the acetic acid to sodium acetate, leaving 0.005 mol acetic acid and 0.005 mol sodium acetate — the definition of a buffer. To reach the equivalence point (all acid converted), an equal volume of NaOH at the same concentration would be required.

Ka is the equilibrium constant for the reaction HA ⇌ H⁺ + A⁻, quantifying the extent of ionization. A larger Ka indicates a stronger acid (more ionization), not a weaker one — this is the most common misconception. Like all equilibrium constants, Ka does change with temperature. Weak acids by definition have Ka values much less than 1; strong acids have very large Ka values.

Methylamine is a weak base: CH₃NH₂ + H₂O ⇌ CH₃NH₃⁺ + OH⁻. From the ICE table, x²/(0.10 − x) = 4.4 × 10⁻⁴. Because x is not negligible (>5%), solving the quadratic gives x ≈ 6.4 × 10⁻³ M = [OH⁻]. pOH = −log(6.4 × 10⁻³) ≈ 2.19, so pH = 14 − 2.19 ≈ 11.8. A pH of 3.2 is acidic and physically impossible for a base solution.

From pH = 3.00, [H⁺] = [X⁻] = 1.0 × 10⁻³ M. The equilibrium concentration of HX is 0.10 − 0.001 = 0.099 M ≈ 0.10 M. Ka = [H⁺][X⁻]/[HX] = (1.0 × 10⁻³)²/0.099 ≈ 1.0 × 10⁻⁵. A common error is to report Ka = 1.0 × 10⁻³ by confusing [H⁺] with Ka directly, which ignores the equilibrium expression.

Using Henderson-Hasselbalch: pH = pKa + log([A⁻]/[HA]) = 4.74 + log(0.40/0.20) = 4.74 + log(2.0) = 4.74 + 0.30 = 5.04. The pH is above pKa because there is more conjugate base than acid. If the concentrations were equal, pH would equal pKa = 4.74. Reversing the ratio gives 4.74 − 0.30 = 4.44.

When OH⁻ is added to a buffer, the weak acid component reacts: HA + OH⁻ → A⁻ + H₂O. This consumes the hydroxide and produces conjugate base, causing only a small change in the [A⁻]/[HA] ratio and a minimal pH shift. The conjugate base (A⁻) does not react with OH⁻ because both are bases. Dilution is negligible for small additions, and strong bases do not become weaker bases in solution.

Percent ionization = ([H⁺]/C₀) × 100%, which increases with Ka at the same initial concentration. HD has the largest Ka (1.0 × 10⁻³), making it the strongest of these four weak acids. At 0.10 M, HD ionizes approximately 9.5%, while HA ionizes only about 0.1%. A larger Ka means the equilibrium lies further toward products, producing more H⁺.

When a strong acid (e.g., HCl) reacts with a strong base (e.g., NaOH), the products are a neutral salt (NaCl) and water. Neither Na⁺ nor Cl⁻ hydrolyzes in water, so the solution remains neutral at pH = 7 at 25°C. This is distinct from a weak acid–strong base titration, where the conjugate base at the equivalence point does hydrolyze, producing a basic solution (pH > 7).

An amphiprotic species can both donate and accept a proton. HCO₃⁻ can donate a proton to form CO₃²⁻ (acting as an acid) or accept a proton to form H₂CO₃ (acting as a base). SO₄²⁻ and Cl⁻ are conjugate bases of strong acids and essentially cannot accept protons in aqueous solution. NH₄⁺ can only donate a proton to form NH₃ and does not meaningfully accept protons.

Using Henderson-Hasselbalch: pH = pKa + log([HPO₄²⁻]/[H₂PO₄⁻]). Substituting: 6.9 = 7.2 + log(ratio), so log(ratio) = −0.3, and ratio = 10^(−0.3) ≈ 0.50. This makes physical sense: since the target pH (6.9) is below pKa (7.2), there must be more of the acid form (H₂PO₄⁻) than the conjugate base (HPO₄²⁻), confirming the ratio must be less than 1.

For a conjugate acid-base pair, Ka × Kb = Kw = 1.0 × 10⁻¹⁴. Therefore Kb = Kw/Ka = (1.0 × 10⁻¹⁴)/(1.8 × 10⁻⁵) = 5.6 × 10⁻¹⁰. This very small Kb confirms that acetate is an extremely weak base. Choice A (1.8 × 10⁻⁵) is a common error where students set Ka equal to Kb. Choice D reverses the division, computing Ka/Kw instead.

Moles HCl = 0.0250 L × 0.100 M = 2.50 × 10⁻³ mol. Moles NaOH = 0.0260 L × 0.100 M = 2.60 × 10⁻³ mol. Excess NaOH = 1.0 × 10⁻⁴ mol. Total volume = 51.0 mL. [OH⁻] = (1.0 × 10⁻⁴ mol)/(0.0510 L) = 1.96 × 10⁻³ M. pOH = −log(1.96 × 10⁻³) = 2.71, so pH = 14 − 2.71 = 11.3. Choice 2.71 is the pOH, not the pH — a very common error when students forget to convert.

Because Ka2 ≪ Ka1, only the first ionization contributes significantly to [H⁺]. ICE table: x²/(0.10 − x) = 1.0 × 10⁻³. The 5% approximation fails here (x/0.10 > 5%), so the quadratic must be solved: x ≈ 9.5 × 10⁻³ M. pH = −log(9.5 × 10⁻³) ≈ 2.0. A pH of 3.0 would result from incorrectly assuming Ka1 applies directly without an ICE table. pH 5.5 would be far too high for an acid at this concentration with Ka1 = 10⁻³.

pKa = −log(1.8 × 10⁻⁴) = 3.74. Adding NaOH converts weak acid to conjugate base: [HA] = 0.150 − 0.020 = 0.130 mol; [A⁻] = 0.100 + 0.020 = 0.120 mol. By Henderson-Hasselbalch: pH = 3.74 + log(0.120/0.130) = 3.74 + log(0.923) = 3.74 − 0.035 = 3.71. A pH of 3.74 (pKa) would only apply if [A⁻] = [HA]. The value 3.57 approximates the initial buffer pH before NaOH was added.

Buffer capacity depends primarily on the total concentration (moles) of buffering components — the more moles of weak acid and conjugate base present, the more strong acid or base the buffer can absorb. Adding equal moles of both directly increases the total buffer concentration. Diluting with water (choice A) reduces concentration, decreasing buffer capacity. NaCl (choice C) provides no buffering species. Replacing sodium acetate with NaCl (choice D) destroys the buffer entirely.

At the half-equivalence point, exactly half the NH₃ has been converted to NH₄⁺, so [NH₃] = [NH₄⁺]. Henderson-Hasselbalch gives pH = pKa(NH₄⁺) when the ratio equals 1. Since pKb(NH₃) = −log(1.8 × 10⁻⁵) = 4.74, and pKa(NH₄⁺) = 14 − pKb = 14 − 4.74 = 9.26, the pH = 9.26. The value 4.74 is the pKb of ammonia, not the pH. A pH of 7.00 would only occur at the equivalence point of a strong acid–strong base titration.

Initial pH = 9.26 + log(0.100/0.100) = 9.26. Adding HCl converts NH₃ to NH₄⁺: [NH₃] = 0.090 mol, [NH₄⁺] = 0.110 mol. New pH = 9.26 + log(0.090/0.110) = 9.26 + log(0.818) = 9.26 − 0.087 ≈ 9.17. The change is 9.17 − 9.26 = −0.09. A decrease of 0.57 would suggest the buffer was overwhelmed, which does not occur here since only 10% of the NH₃ was consumed.

At 5.0% ionization, [H⁺] = 0.050 × C₀. At equilibrium, [HA] = C₀ − 0.050C₀ = 0.95C₀. Substituting into Ka: Ka = (0.050C₀)²/(0.95C₀) = 0.0025C₀/0.95 = 0.00263C₀. Solving: C₀ = Ka/0.00263 = (1.0 × 10⁻⁵)/0.00263 = 3.8 × 10⁻³ M. This problem illustrates that percent ionization increases as concentration decreases — a more dilute solution of the same acid would show higher percent ionization.

CO₃²⁻ is the conjugate base of HCO₃⁻ and hydrolyzes: CO₃²⁻ + H₂O ⇌ HCO₃⁻ + OH⁻. First, Kb = Kw/Ka2 = (1.0 × 10⁻¹⁴)/(4.7 × 10⁻¹¹) = 2.1 × 10⁻⁴. Solving x²/(0.10 − x) = 2.1 × 10⁻⁴ by quadratic gives x ≈ 4.5 × 10⁻³ M = [OH⁻]. pOH = −log(4.5 × 10⁻³) ≈ 2.35, so pH = 14 − 2.35 ≈ 11.7. A pH of 9.3 is a common error from incorrectly using Ka2 directly instead of computing Kb from the conjugate base relationship.

HNO₃ is one of the seven strong acids and dissociates completely in water: HNO₃ → H⁺ + NO₃⁻. HF, CH₃COOH, and HCN are all weak acids that only partially ionize in solution, each with a Ka much less than 1.

NaOH is a strong base and dissociates completely, so [OH⁻] = 0.010 M. pOH = −log(0.010) = 2. pH = 14 − pOH = 14 − 2 = 12. A common error is reporting pOH as the pH, which would give the incorrect answer of 2.

A conjugate acid is formed when a base accepts a proton (H⁺). NH₃ acts as a base by accepting a proton from water to form NH₄⁺, which is its conjugate acid. H₂O is the acid in this reaction, and OH⁻ is the conjugate base of H₂O — not of NH₃.

[H⁺] = 10^(−pH) = 10^(−3) = 1.0 × 10⁻³ M. Choice D (1.0 × 10⁻¹¹ M) represents [OH⁻], since pOH = 14 − 3 = 11. Choice C incorrectly uses the pH value as a coefficient rather than as a negative exponent.

For any conjugate acid-base pair, Ka × Kb = Kw = 1.0 × 10⁻¹⁴ at 25°C. This allows calculation of Kb when Ka is known, and vice versa. Choice A confuses this with pH + pOH = 14, which relates concentrations of H⁺ and OH⁻, not equilibrium constants.

A Brønsted-Lowry acid is defined as a proton (H⁺) donor. Choice D describes an Arrhenius base. Choices A and B describe a Lewis acid and a Lewis base, respectively. The Brønsted-Lowry definition is broader than the Arrhenius definition and applies in non-aqueous contexts as well.

In pure water, [H⁺] = [OH⁻] by the autoionization equilibrium H₂O ⇌ H⁺ + OH⁻. Setting [H⁺] = [OH⁻] = x: x² = 1.0 × 10⁻¹⁴, so x = 1.0 × 10⁻⁷ M, giving pH = 7.0 (neutral at 25°C). Choice A incorrectly uses Kw itself as the concentration rather than its square root.

Using the Henderson-Hasselbalch equation: pH = pKa + log([A⁻]/[HA]) = −log(1.8 × 10⁻⁵) + log(0.20/0.10) = 4.74 + log(2) = 4.74 + 0.30 = 5.04. Because [A⁻] > [HA], the pH is above the pKa. Choice B (4.74) would be correct only if the moles of acid and conjugate base were equal.

At pH 3.00, [H⁺] = 10⁻³ = 1.0 × 10⁻³ M. Each ionized HA molecule produces one H⁺, so the ionized concentration is 1.0 × 10⁻³ M. Percent ionization = (1.0 × 10⁻³ / 0.100) × 100% = 1.0%. Choice D (10%) would result from a pH of 2.00. This confirms HA is a weak acid, since strong acids ionize 100%.

NH₄Cl produces an acidic solution because NH₄⁺ is the conjugate acid of the weak base NH₃ and undergoes hydrolysis: NH₄⁺ + H₂O ⇌ NH₃ + H₃O⁺. NaCl is neutral because Na⁺ and Cl⁻ are conjugates of a strong base and strong acid, respectively. NaF and NaCH₃COO produce basic solutions because F⁻ and CH₃COO⁻ are conjugate bases of weak acids that hydrolyze to produce OH⁻.

At the half-equivalence point, exactly half the weak acid has been neutralized, so [HA] = [A⁻]. Substituting into Henderson-Hasselbalch: pH = pKa + log([A⁻]/[HA]) = pKa + log(1) = pKa. This key relationship is used experimentally to read Ka directly from the inflection region of a titration curve.

At the equivalence point, all acetic acid has been converted to sodium acetate. The acetate ion is the conjugate base of a weak acid and hydrolyzes: CH₃COO⁻ + H₂O ⇌ CH₃COOH + OH⁻, producing a basic solution with pH greater than 7. A pH of 7.00 at the equivalence point occurs only for strong acid-strong base titrations. Choice D describes the half-equivalence point, not the equivalence point.

When a weak acid is diluted, the equilibrium HA ⇌ H⁺ + A⁻ shifts right by Le Chatelier's principle, increasing the fraction of HA that ionizes. While the absolute [H⁺] decreases upon dilution, percent ionization = ([H⁺] / [HA]initial) × 100% rises. Ka remains constant with dilution (temperature is unchanged), but this constant Ka means a higher ionization fraction at lower concentrations.

After the first ionization, H₂SO₃ → H⁺ + HSO₃⁻, the resulting anion carries a negative charge. Removing a second proton from HSO₃⁻ requires working against the electrostatic attraction between the departing H⁺ and the negatively charged ion, making Ka2 far smaller than Ka1. This pattern — each successive Ka being orders of magnitude smaller — holds universally for polyprotic acids. Dilution does not affect equilibrium constants.

For the equilibrium C₅H₅N + H₂O ⇌ C₅H₅NH⁺ + OH⁻, [OH⁻] = x ≈ √(Kb × C) = √(1.7 × 10⁻⁹ × 0.10) = √(1.7 × 10⁻¹⁰) ≈ 1.3 × 10⁻⁵ M. The approximation x << 0.10 is valid because Kb is very small. Choice A incorrectly equates Kb directly with [OH⁻]. Choice D incorrectly takes Kb × C without applying the square root.

Buffer capacity is determined by the moles of weak acid and conjugate base available to neutralize added acid or base. Buffer X has 10 times higher concentrations, so it can absorb far more added H⁺ or OH⁻ before pH shifts significantly. Both buffers share the same pH and the same [A⁻]/[HA] ratio, but their absolute amounts differ by a factor of 10. The pKa alone does not determine buffer capacity.

An indicator displays a visible mixture of both colors when the pH is within approximately one unit of its pKa. Since pKa = 5.0, the transition range is approximately pH 4 to pH 6. Below pH 4, essentially all indicator is in the HIn form (red); above pH 6, it is predominantly In⁻ (yellow). Choosing an indicator whose transition range brackets the equivalence point pH is essential for accurate titration endpoint detection.

For any conjugate acid-base pair, Ka × Kb = Kw = 1.0 × 10⁻¹⁴, so Kb = Kw / Ka. HD has the smallest Ka (1.0 × 10⁻⁹), giving its conjugate base D⁻ the largest Kb = 1.0 × 10⁻¹⁴ / 1.0 × 10⁻⁹ = 1.0 × 10⁻⁵. The weaker the acid, the stronger its conjugate base. HA, as the strongest acid in this set, has the weakest conjugate base (Kb = 1.0 × 10⁻¹¹).

Moles of CH₃COOH = 0.0500 L × 0.200 M = 0.0100 mol. Moles of NaOH = 0.0250 L × 0.200 M = 0.00500 mol. After reaction: 0.00500 mol CH₃COOH remains and 0.00500 mol CH₃COO⁻ is produced. Since [CH₃COOH] = [CH₃COO⁻], Henderson-Hasselbalch gives pH = pKa + log(1) = pKa = −log(1.8 × 10⁻⁵) = 4.74. This corresponds to the half-equivalence point. The volume change does not affect pH because it cancels in the concentration ratio.

Moles CH₃NH₂ = 0.0300 L × 0.200 M = 0.00600 mol. Moles HCl = 0.0200 L × 0.200 M = 0.00400 mol. After reaction: CH₃NH₂ = 0.00200 mol, CH₃NH₃⁺ = 0.00400 mol. pKa of CH₃NH₃⁺ = 14 − pKb = 14 − (−log(4.4 × 10⁻⁴)) = 14 − 3.36 = 10.64. pH = pKa + log([CH₃NH₂]/[CH₃NH₃⁺]) = 10.64 + log(0.00200/0.00400) = 10.64 + log(0.5) = 10.64 − 0.30 = 10.34. Choice B (10.64) is the pH only at the half-equivalence point where [CH₃NH₂] = [CH₃NH₃⁺].

[H⁺] = 10^(−4.15) ≈ 7.1 × 10⁻⁵ M. At equilibrium, [H⁺] = [A⁻] = 7.1 × 10⁻⁵ M and [HA] = 0.050 − 7.1 × 10⁻⁵ ≈ 0.04993 M. Ka = [H⁺][A⁻] / [HA] = (7.1 × 10⁻⁵)² / 0.04993 ≈ 5.0 × 10⁻⁹ / 0.04993 ≈ 1.0 × 10⁻⁷. Choice B incorrectly uses [H⁺] as Ka without squaring and dividing. Failing to subtract x from the initial concentration (or ignoring the denominator) leads to choices C or D.

Adding CH₃COONa introduces additional CH₃COO⁻ (the common ion). By Le Chatelier's principle, the equilibrium CH₃COOH ⇌ H⁺ + CH₃COO⁻ shifts left, reducing [H⁺] and raising pH. Because the shift also leaves more CH₃COOH un-ionized, the fraction that ionizes — percent ionization — decreases as well. Ka remains constant since temperature is unchanged. Both effects are direct consequences of the common ion effect.

HPO₄²⁻ is amphiprotic: as an acid it can donate a proton (Ka3 = 4.8 × 10⁻¹³), and as a base it can accept one (Kb = Kw/Ka2 = 1.0 × 10⁻¹⁴ / 6.2 × 10⁻⁸ = 1.6 × 10⁻⁷). Since Kb (1.6 × 10⁻⁷) >> Ka3 (4.8 × 10⁻¹³), the base behavior dominates by six orders of magnitude, making the solution basic. Na⁺ is the conjugate cation of a strong base and does not hydrolyze.

At the equivalence point, all HA is converted to A⁻ (0.0500 M in 50.0 mL total). Kb of A⁻ = Kw/Ka = 1.0 × 10⁻¹⁴ / 1.0 × 10⁻⁶ = 1.0 × 10⁻⁸. [OH⁻] = √(Kb × C) = √(1.0 × 10⁻⁸ × 0.0500) ≈ 2.2 × 10⁻⁵ M, giving pOH ≈ 4.66 and pH ≈ 9.34. Phenolphthalein, which changes color across pH 8.2–10.0, brackets this value. All other listed indicators transition at pH values far below 9.34 and would indicate a false endpoint.

pKa = −log(1.4 × 10⁻⁴) = 3.85. Adding 0.020 mol NaOH converts 0.020 mol HA to 0.020 mol A⁻. New amounts: HA = 0.100 − 0.020 = 0.080 mol; A⁻ = 0.100 + 0.020 = 0.120 mol. pH = pKa + log([A⁻]/[HA]) = 3.85 + log(0.120/0.080) = 3.85 + log(1.5) = 3.85 + 0.18 = 4.03. The buffer resists large pH changes — the pH rises by only 0.18 units despite the addition of strong base. Choice A (3.85) is the pH of the original buffer before NaOH is added.

The Brønsted-Lowry definition of a base is a species that accepts a proton (H+). Choice A describes a Brønsted-Lowry acid. Choice C describes a Lewis base, which is a broader concept that does not require proton transfer. Choice D describes an Arrhenius base, which is limited to aqueous solutions and requires production of OH-.

A conjugate acid-base pair differs by exactly one proton. H3O+ and H2O differ by one proton — H3O+ donates H+ to become H2O — making them a conjugate pair. HCl and NaOH are not related by proton transfer at all. H2SO4 and SO4²⁻ differ by two protons, so they are not a conjugate pair. NH3 and OH⁻ are not related by a single proton transfer.

Nitric acid (HNO3) is one of the six common strong acids and dissociates essentially completely in aqueous solution. HF, despite containing a halogen, is a weak acid with Ka = 7.2 × 10⁻⁴. Acetic acid and phosphoric acid are also weak acids with Ka values far less than 1. A Ka value listed in the answer choices is itself a signal that the acid is weak, since strong acids have no meaningful Ka to report.

HCl is a strong acid that dissociates completely in water, so [H+] = 0.0010 M = 1.0 × 10⁻³ M. pH = -log(1.0 × 10⁻³) = 3. Choice B (pH = 2) corresponds to [H+] = 0.010 M, and Choice A (pH = 1) corresponds to [H+] = 0.10 M — each represents a concentration ten times higher than given.

For any conjugate acid-base pair, Ka × Kb = Kw = 1.0 × 10⁻¹⁴ at 25°C. This follows from combining the two equilibria: HA ⇌ H+ + A⁻ (Ka) and A⁻ + H2O ⇌ HA + OH⁻ (Kb). Adding those two reactions gives the net autoionization of water (Kw), and multiplying the individual constants gives Kw. This relationship is used to calculate Kb when Ka is known, or vice versa.

An amphoteric substance can act as either an acid or a base. Water is amphoteric because it can donate a proton (acting as a Brønsted-Lowry acid, as in H2O + NH3 → OH⁻ + NH4+) or accept a proton (acting as a Brønsted-Lowry base, as in H2O + HCl → H3O+ + Cl⁻). Choice C describes amphiprotic character in a redox sense, which is a separate concept. Water does not fully dissociate (Choice A) and does not form a buffer with strong acids or bases (Choice D).

At 25°C, pH + pOH = 14. Therefore pH = 14 − 9.0 = 5.0. Since pH = 5.0 is less than 7.0, the solution is acidic. This makes physical sense because pOH = 9.0 means [OH⁻] = 10⁻⁹ M, which is less than the neutral value of 10⁻⁷ M, confirming excess H+. Choice A incorrectly sets pH equal to pOH, and Choice C correctly calculates the pH but misidentifies the classification.

In pure water the autoionization produces equal concentrations of H3O+ and OH⁻. Setting [H3O+] = [OH⁻] = x and using Kw = x² = 1.0 × 10⁻¹⁴ gives x = 1.0 × 10⁻⁷ M. Choices B and D are physically unrealistic because setting either ion to 1.0 M would mean pure water behaves like a concentrated strong acid or base. Choice C uses the correct [H3O+] but sets [OH⁻] equal to Kw itself rather than its square root.

Using the Henderson-Hasselbalch equation: pH = pKa + log([A⁻]/[HA]). Here pKa = −log(1.8 × 10⁻⁴) = 3.74. pH = 3.74 + log(0.30/0.20) = 3.74 + log(1.5) = 3.74 + 0.18 = 3.92. Choice B (3.74) is the pKa itself, which applies only when [HA] = [A⁻]. Choice A results from subtracting the log term instead of adding it.

Adding sodium acetate introduces acetate ions (A⁻), the common ion. By Le Chatelier's principle, the equilibrium CH3COOH ⇌ H+ + CH3COO⁻ shifts to the left, suppressing further ionization of acetic acid (degree of ionization decreases). This leftward shift reduces [H+], so the pH rises. Choice C is self-contradictory: if ionization increased, [H+] would rise and pH would fall, not rise.

At the equivalence point, all acetic acid has been converted to sodium acetate. The acetate ion is the conjugate base of a weak acid and undergoes hydrolysis: CH3COO⁻ + H2O ⇌ CH3COOH + OH⁻, producing OH⁻ and making the solution basic (pH > 7). Choice B is wrong because combining equal moles of a weak acid and a strong base yields a solution of the conjugate base, which is itself basic. Na+ is a spectator ion with no effect on pH.

When a weak acid is diluted, Le Chatelier's principle predicts that the equilibrium HA ⇌ H+ + A⁻ shifts to the right to partially counteract the decrease in concentration of all species. Although [H+] falls in absolute terms, the fraction of HA that has ionized increases. Ka depends only on temperature and remains constant (making Choice A wrong). Choice B confuses a constant Ka with a constant degree of ionization — these are not the same thing.

Acid strength increases with increasing Ka. Ordering by Ka value: HCN (6.2 × 10⁻¹⁰) < CH3COOH (1.8 × 10⁻⁵) < HCOOH (1.8 × 10⁻⁴) < HF (7.2 × 10⁻⁴). Choice D incorrectly reverses CH3COOH and HCOOH — formic acid (Ka = 1.8 × 10⁻⁴) is about ten times stronger than acetic acid (Ka = 1.8 × 10⁻⁵). Choice B reverses the entire ranking.

Buffer capacity depends on the total moles of weak acid and conjugate base available to neutralize added strong acid or base, not just their ratio. Buffer B contains 10 times as many moles of each component as Buffer A and can therefore absorb far more added H+ or OH⁻ before the pH changes significantly. Choice C incorrectly equates the same pH ratio with the same buffer capacity — capacity is determined by concentration, not by ratio alone.

A suitable indicator must change color within the steep portion of the titration curve near the equivalence point. Because the equivalence point occurs at pH ≈ 9.0, phenolphthalein (transition range pH 8.2 to 10.0) is the best choice — its color change brackets the equivalence point. Methyl orange and bromocresol green both change color at pH values far below the equivalence point. Bromothymol blue transitions in the neutral region (pH 6–7.6) and would signal a false endpoint well before the true equivalence point.

Setting up the base-ionization equilibrium: Kb = x²/(0.050 − x) ≈ x²/0.050 = 1.8 × 10⁻⁵. Solving: x = [OH⁻] = √(9.0 × 10⁻⁷) ≈ 9.5 × 10⁻⁴ M. Then pOH = −log(9.5 × 10⁻⁴) ≈ 3.02, and pH = 14 − 3.02 ≈ 11.0. Choice A (pH 3.0) would describe an acidic solution with pOH = 11. Choice D (pH 12.5) would require a much stronger or more concentrated base.

NH4Cl is acidic because NH4+ is the conjugate acid of the weak base NH3 and undergoes hydrolysis: NH4+ ⇌ NH3 + H+, releasing H+ and lowering the pH below 7. NaCl is neutral because both ions come from a strong acid and a strong base and do not hydrolyze. NaCH3COO is basic because CH3COO⁻ is the conjugate base of a weak acid. Na2CO3 is also basic because CO3²⁻ is a moderately strong base.

pKa = −log(6.3 × 10⁻⁵) = 4.20. The added HCl reacts with the conjugate base: C6H5COO⁻ + H+ → C6H5COOH. After reaction: mol acid = 0.20 + 0.050 = 0.25 mol and mol conjugate base = 0.30 − 0.050 = 0.25 mol. Applying Henderson-Hasselbalch: pH = 4.20 + log(0.25/0.25) = 4.20 + 0 = 4.20. The ratio [A⁻]/[HA] becomes exactly 1 after addition, so the log term vanishes. Choice D (4.38) was the initial buffer pH before any HCl was added.

At the half-equivalence point, exactly half the original acid has been neutralized, so moles of HA remaining equal moles of A⁻ formed: [HA] = [A⁻]. Substituting into the Henderson-Hasselbalch equation: pH = pKa + log([A⁻]/[HA]) = pKa + log(1) = pKa = −log(1.0 × 10⁻⁶) = 6.00. Choice A is incorrect because pKa = 6.00, not 7.00 — the half-equivalence point equals pKa regardless of whether that value is above or below 7. Choice D describes a formula for a different situation entirely.

For a polyprotic system, each species predominates between its flanking pKa values. H2Gly+ (fully protonated cation) predominates below pKa1 = 2.35. The neutral zwitterion HGly predominates between pKa1 and pKa2, that is, from pH 2.35 to pH 9.77. Gly⁻ (deprotonated anion) predominates above pKa2 = 9.77. Choice D is incorrect: the isoelectric point (pH ≈ 6.06) is the pH of minimum solubility and zero net charge, but HGly predominates across the entire range from 2.35 to 9.77, not only at one specific pH.

Moles of HF = 0.0200 L × 0.100 mol/L = 2.00 mmol. Moles of NaOH added = 0.0220 L × 0.100 mol/L = 2.20 mmol. NaOH neutralizes all 2.00 mmol HF, leaving 0.20 mmol NaOH in excess. Total volume = 42.0 mL. [OH⁻]excess = 0.20 mmol / 42.0 mL = 4.76 × 10⁻³ M. Because Kb(F⁻) = Kw/Ka = 1.4 × 10⁻¹¹ is negligible relative to the excess OH⁻, pOH = −log(4.76 × 10⁻³) = 2.32 and pH = 14 − 2.32 = 11.68. Choice A results from incorrectly halving the excess NaOH concentration before calculating pOH.

At the equivalence point, acetic acid and ammonia react to form ammonium acetate. The pH of such a solution is given by pH = 7 + (pKa − pKb)/2. Here pKa(CH3COOH) = 4.74 and pKb(NH3) = 4.74, so pH = 7 + (4.74 − 4.74)/2 = 7.00. Equivalently, Ka(CH3COOH) = Kb(NH3), meaning the hydrolysis of CH3COO⁻ and the hydrolysis of NH4+ proceed at identical rates, producing equal amounts of OH⁻ and H+, which cancel. Choice C would apply only if the solution contained NH4+ with no competing acetate hydrolysis.

At pH = 3.00, [H+] = [A⁻] = 1.00 × 10⁻³ M. The equilibrium concentration of HA is 0.200 − 0.001 = 0.199 M. Ka = [H+][A⁻] / [HA] = (1.00 × 10⁻³)² / 0.199 = 1.00 × 10⁻⁶ / 0.199 ≈ 5.0 × 10⁻⁶. Choice B (5.0 × 10⁻³) results from failing to square [H+] in the numerator — using [H+] × 0.200 rather than [H+]² / [HA]. Choice D (1.0 × 10⁻⁶) results from using 0.200 M as the denominator without subtracting the amount ionized and without properly squaring the numerator.

In BF3 + F⁻ → BF4⁻, fluoride ion donates an electron pair (Lewis base) to the boron atom of BF3, which accepts it (Lewis acid), forming a new coordinate covalent bond with no proton transferred. This is a purely Lewis acid-base reaction. Choices A, C, and D all involve transfer of H+ from a donor to an acceptor, qualifying them as Brønsted-Lowry reactions. BF3 is the textbook example of a Lewis acid: boron has only six valence electrons and an empty orbital available to accept an electron pair.

In a neutral solution [H3O+] = [OH⁻] at any temperature. At 37°C: [H3O+] = √(2.4 × 10⁻¹⁴) = 1.55 × 10⁻⁷ M, giving pH = −log(1.55 × 10⁻⁷) = 6.81. This solution is still neutral because both ions are present in equal concentrations. A solution is acidic only when [H3O+] > [OH⁻]. The increased Kw reflects greater autoionization at higher temperature, but both H3O+ and OH⁻ increase equally, preserving neutrality. Choice A is incorrect: pH = 7.00 corresponds to neutrality only at 25°C, where Kw = 1.0 × 10⁻¹⁴.

The Bronsted-Lowry definition defines an acid as a proton (H+) donor. Choice A describes a Bronsted-Lowry base. Choice C describes a Lewis base. Choice D describes a substance that neutralizes acid by releasing OH-, but this is not the Bronsted-Lowry definition of an acid.

HNO3 is one of the seven strong acids that dissociate completely in aqueous solution. HF is a weak acid (Ka = 7.2 x 10^-4) despite containing a halogen, because the H-F bond is particularly strong. HCN and CH3COOH are both weak acids with small Ka values.

HCl is a strong acid and dissociates completely, so [H+] = 0.010 M = 1.0 x 10^-2 M. pH = -log(1.0 x 10^-2) = 2. Choice A (pH = 1) would correspond to 0.10 M HCl. Choices C and D correspond to basic solutions, not acidic ones.

A conjugate base is formed when an acid donates one proton. H2PO4^- loses one H+ to form HPO4^2-, which is its conjugate base. H3PO4 is the conjugate acid of H2PO4^- (formed by gaining a proton). PO4^3- is two deprotonation steps away from H2PO4^-, not one.

At 25 degrees C, pH + pOH = 14 (derived from Kw = 1.0 x 10^-14). Therefore pH = 14 - pOH = 14 - 3.0 = 11.0. This solution is basic. A pOH of 3.0 means [OH-] = 1.0 x 10^-3 M, which exceeds [H+] and confirms a basic solution.

HCO3^- (bicarbonate) is amphoteric: it can donate a proton (acting as an acid) to form CO3^2-, or accept a proton (acting as a base) to form H2CO3. Cl^- is only a negligibly weak base because it is the conjugate base of the strong acid HCl. HNO3 only donates protons. Na^+ does not participate in acid-base reactions in aqueous solution.

A buffer resists pH changes when small amounts of acid or base are added, because it contains both a weak acid and its conjugate base. It does not maintain a fixed pH of 7 — buffer pH is determined by the pKa of the weak acid and the ratio of components. Buffers have limited capacity and cannot neutralize unlimited amounts of added acid or base. Buffers are made from weak acids (not strong acids) and their conjugate bases.

Using the approximation \([H^+] = \sqrt{K_a \times C}\): \([H^+] = \sqrt{1.0 \times 10^{-4} \times 0.10} = \sqrt{1.0 \times 10^{-5}} = 3.16 \times 10^{-3}\) M. \(pH = -\log(3.16 \times 10^{-3}) = 2.50\). The approximation is valid because percent ionization \(= 3.16 \times 10^{-3} / 0.10 \times 100 = 3.2\%\), which is under \(5\%\). Choice B (\(pH = 3.0\)) would require \([H^+] = 1.0 \times 10^{-3}\) M, which is inconsistent with this \(K_a\) and concentration.

Using the Henderson-Hasselbalch equation: pH = pKa + log([A-]/[HA]) = 4.74 + log(0.10/0.30) = 4.74 + log(0.333) = 4.74 - 0.477 = 4.26. Since there is more acid than conjugate base, the pH falls below pKa. Choice B (4.74) is the pH only when [A-] = [HA], which is not the case here.

Percent ionization = ([H+] / [HA]initial) x 100. From pH = 3.15, [H+] = 10^(-3.15) = 7.1 x 10^-4 M. Percent ionization = (7.1 x 10^-4 / 0.20) x 100 = 0.35%. Choice C (3.5%) is off by a factor of 10, a common error from misplacing a decimal. This relatively low percent ionization is consistent with Le Chatelier's principle: higher initial concentration suppresses dissociation.

NH4Cl dissociates to give NH4+ and Cl^-. NH4+ is the conjugate acid of the weak base NH3 and undergoes hydrolysis: NH4+ + H2O = NH3 + H3O+, producing an acidic solution. NaCl is neutral because both Na+ and Cl^- come from a strong base and strong acid respectively. NaF is basic because F^- (conjugate base of the weak acid HF) accepts protons from water. Na2CO3 is basic because CO3^2- is the conjugate base of the weak acid HCO3^-.

When HCl is added, it reacts with NH3: NH3 + HCl → NH4+. This consumes 0.10 mol of NH3 and produces 0.10 mol of NH4+. New amounts: NH3 = 0.40 mol, NH4+ = 0.60 mol. Using Henderson-Hasselbalch: pH = 9.26 + log(0.40/0.60) = 9.26 + log(0.667) = 9.26 - 0.176 = 9.08. Choice B (9.26) was the original pH when equal moles of NH3 and NH4+ were present; adding acid shifts the ratio and lowers the pH.

For a strong acid-strong base titration, the equivalence point pH is always exactly 7.00 because neither product ion hydrolyzes water. A pH above 7 at the equivalence point indicates the conjugate base of a weak acid is present and produces OH^- by hydrolyzing water. Equal volumes being needed simply means the acid and base concentrations are equal — this is consistent with either a strong or weak acid and does not distinguish between them.

Acid strength increases directly with Ka. Ranking numerically: CH3COOH (1.8 x 10^-5) < HNO2 (4.5 x 10^-4) < HF (7.2 x 10^-4). A larger Ka means greater extent of dissociation and stronger acid. Choice D is incorrect because it places HF before HNO2, but Ka of HF (7.2 x 10^-4) is larger than Ka of HNO2 (4.5 x 10^-4), making HF the strongest of the three.

At the half-equivalence point, exactly half of the weak acid has been converted to its conjugate base, so [HA] = [A^-]. The Henderson-Hasselbalch equation becomes pH = pKa + log(1) = pKa. Therefore pKa = 4.20, and Ka = 10^(-4.20) = 6.3 x 10^-5. Choice A confuses pKa with Ka — they are related by Ka = 10^(-pKa) and have very different numerical values. The half-equivalence point is a key landmark on a titration curve precisely because it directly reveals pKa.

Effective buffers are prepared using acid-base pairs whose pKa is within approximately 1 pH unit of the target pH. The target pH is 9.2, and the pKa of NH4+/NH3 is 9.26, which is the closest match. Acetic acid (pKa 4.74) is more than 4 units away and would provide negligible buffering at pH 9.2. Carbonic acid (pKa1 6.35) and phosphate (pKa2 7.20) are also too far from the target.

There is an inverse relationship between acid strength and conjugate base strength: the weaker the acid, the stronger its conjugate base. HCN (Ka = 6.2 x 10^-10) is far weaker than HF (Ka = 7.2 x 10^-4), so CN^- is a stronger base than F^-. This is confirmed by Kb values: Kb(CN^-) = Kw/Ka(HCN) = 1.6 x 10^-5, while Kb(F^-) = Kw/Ka(HF) = 1.4 x 10^-11. Choice A incorrectly states that a larger Ka for the parent acid leads to a stronger conjugate base, which is backwards.

The Henderson-Hasselbalch equation, pH = pKa + log([A^-]/[HA]), shows that pH depends on the ratio of conjugate base to acid, not their absolute concentrations. When a buffer is diluted with water, both components are diluted by the same factor, so their ratio remains unchanged and pH is essentially constant. Choice D is incorrect — pure water does not neutralize or buffer; it simply reduces the concentration of both components proportionally.

At the equivalence point, all acetic acid is converted to acetate ion. Total volume \(= 100.0\) mL, moles of acetate \(= 5.00\) mmol, so $[CH_3COO^-] = 0.0500$ M. Acetate undergoes hydrolysis: \(K_b = K_w/K_a = 1.0 \times 10^{-14} / 1.8 \times 10^{-5} = 5.56 \times 10^{-10}\). \([OH^-] = \sqrt{5.56 \times 10^{-10} \times 0.0500} = 5.27 \times 10^{-6}\) M. $pOH = 5.28$, \(pH = 14 - 5.28 = 8.72\). Choice A (pH \(7.00\)) is wrong because the conjugate base of a weak acid hydrolyzes water to produce excess \(OH^-\), making the equivalence point of any weak acid-strong base titration basic.

From pH = 2.70: [H+] = 10^(-2.70) = 2.0 x 10^-3 M. At equilibrium: [H+] = [A^-] = 2.0 x 10^-3 M, and [HA] = 0.200 - 0.0020 = 0.198 M. Ka = [H+][A^-] / [HA] = (2.0 x 10^-3)^2 / 0.198 = 4.0 x 10^-6 / 0.198 = 2.0 x 10^-5. A common error is forgetting to subtract [H+] from the initial concentration before dividing (Choice C), which overestimates Ka by approximately tenfold.

An indicator is suitable when its color transition range overlaps with the sharp pH jump at the equivalence point. The equivalence point occurs at pH 8.8, so phenolphthalein (transition range 8.2-10.0) is the correct choice because it will change color at the equivalence point. Methyl orange and bromocresol green both transition at acidic pH values far below 8.8 and would change color long before the equivalence point is reached. Alizarin yellow R transitions above pH 10, well past the equivalence point.

Using Henderson-Hasselbalch: pH = pKa + log([A^-]/[HA]). Substituting: 5.00 = 4.74 + log([A^-]/[HA]). Solving: log([A^-]/[HA]) = 0.26, so [A^-]/[HA] = 10^0.26 = 1.82. More conjugate base than acid is needed because the target pH (5.00) is above pKa (4.74). Choice A (0.55) is the inverse ratio and would instead give a pH below pKa. Choice B (1.00) is the ratio when pH equals pKa exactly.

H3PO4 is a triprotic acid, contributing 3 moles of H+ per mole of acid. Moles of H3PO4 = 0.0300 L x 0.200 mol/L = 0.00600 mol. Total moles of H+ to be neutralized = 3 x 0.00600 = 0.0180 mol. Volume of NaOH = 0.0180 mol / 0.150 mol/L = 0.120 L = 120.0 mL. Choice C (60.0 mL) is a common error that treats H3PO4 as monoprotic and only accounts for one equivalent of NaOH per mole of acid.

Buffer capacity depends on the absolute moles of buffering components. For Buffer X after adding HCl: [CH3COO^-] = 0.090 M, [CH3COOH] = 0.110 M, pH = 4.74 + log(0.090/0.110) = 4.65, a change of 0.09 units. For Buffer Y: [CH3COO^-] = 0.990 M, [CH3COOH] = 1.010 M, pH = 4.74 + log(0.990/1.010) = 4.73, a change of only 0.01 units. Higher concentration buffers resist pH change far more effectively. Choice C is incorrect because while pKa is identical, the differing concentrations produce very different resistance to pH change.

For a neutral solution, \([H^+] = [OH^-] = \sqrt{K_w} = \sqrt{9.6 \times 10^{-14}} = 3.10 \times 10^{-7}\) M. \(pH = -\log(3.10 \times 10^{-7}) = 6.51\). Although this pH is numerically below \(7\), the solution is still neutral because \([H^+]\) equals \([OH^-]\). At \(60\) degrees C, neutrality occurs at pH \(6.51\), not \(7.00\). A solution at pH \(7.00\) at this temperature would actually be slightly basic. This illustrates that neutrality is defined by equal ion concentrations, not by a fixed pH value of \(7\).

HBr is a strong acid that dissociates completely in water, so [H+] = 0.050 M. pH = -log(0.050) = -log(5.0 × 10^-2) = 2 - log(5.0) ≈ 2 - 0.70 = 1.30. Choice B (2.00) would be the pH if [H+] were 0.010 M, a common error from misreading the concentration exponent.

HF (hydrofluoric acid) is a weak acid with Ka = 7.2 × 10^-4; it only partially dissociates in water. HCl, HBr, and HI are all strong hydrohalic acids that dissociate completely. The unusually strong H-F bond, relative to the larger halides, limits dissociation and makes HF the only weak acid among the hydrohalic acids.

A conjugate base is formed by removing exactly one proton from an acid. H2SO4 loses one H+ to become HSO4^- (the hydrogen sulfate ion). SO4^2- is incorrect because it requires removing two protons. H3SO4^+ is the conjugate acid (proton added), not the conjugate base. S^2- results from removing all protons, which does not describe a conjugate base relationship.

From pH = 2.50, [H+] = [A^-] = 10^-2.50 = 3.16 × 10^-3 M. The equilibrium concentration of HA is 0.250 - 0.00316 = 0.247 M. Ka = [H+][A^-]/[HA] = (3.16 × 10^-3)^2 / 0.247 = 9.99 × 10^-6 / 0.247 ≈ 4.0 × 10^-5. Choice B (4.0 × 10^-6) results from incorrectly rounding [H+] to 1.0 × 10^-3 (treating pH as exactly 3.00). Choice C arises from forgetting to square [H+].

The Henderson-Hasselbalch equation is pH = pKa + log([A^-]/[HA]). When the solution is diluted with equal volume of water, both the acid and conjugate base concentrations are halved by the same factor. The ratio [CH3COO^-]/[CH3COOH] remains equal to 1, so log(1) = 0. Therefore pH = 4.74 + 0 = 4.74, unchanged. Dilution does not shift pH as long as the ratio of buffer components is preserved. (Buffer capacity decreases with dilution, but the pH does not change unless dilution is extreme.)

For a conjugate acid-base pair, Ka × Kb = Kw = 1.0 × 10^-14 at 25°C. Therefore Kb = Kw / Ka = (1.0 × 10^-14) / (1.8 × 10^-5) = 5.6 × 10^-10. Choice A (1.8 × 10^-9) results from dividing 1.0 × 10^-14 by a value 10 times too small. Choice C (1.8 × 10^-19) incorrectly multiplies Ka by Kw rather than dividing. This relationship confirms that the weaker an acid, the stronger its conjugate base.

Ka2 (6.3 × 10^-8) is approximately 10^5 times smaller than Ka1 (7.1 × 10^-3). This means the second dissociation produces roughly 100,000 times fewer H+ ions per mole than the first, making its contribution truly negligible. The [H+] established by the first dissociation also suppresses the second step by the common-ion effect, further reducing its contribution. Choices B, C, and D are factually incorrect: the second step is not complete, polyprotic acids do dissociate in multiple steps, and HPO4^2- is soluble.

Added HCl reacts completely with the conjugate base: HCOO^- + H^+ → HCOOH. After the reaction: HCOOH = 0.300 + 0.050 = 0.350 mol; HCOO^- = 0.200 - 0.050 = 0.150 mol. Using Henderson-Hasselbalch: pH = 3.74 + log(0.150/0.350) = 3.74 + log(0.4286) = 3.74 - 0.37 = 3.37. Choice C (3.74) is the initial pH before adding HCl. Choice D (4.11) results from adding the log term instead of subtracting it, reversing the direction of pH change.

At the equivalence point, all benzoic acid is converted to sodium benzoate. Moles of benzoate \(= 0.0500\) L \(\times 0.200\) mol/L \(= 0.0100\) mol. Total volume \(= 100.0\) mL \(= 0.100\) L, so $[C_6H_5COO^-] = 0.100$ M. \(K_b = K_w / K_a = (1.0 \times 10^{-14}) / (6.3 \times 10^{-5}) = 1.59 \times 10^{-10}\). \([OH^-] = \sqrt{K_b \times C} = \sqrt{1.59 \times 10^{-11}} = 3.99 \times 10^{-6}\) M. $pOH = 5.40$; \(pH = 14.00 - 5.40 = 8.60\). The pH is above \(7\) because benzoate is a weak base. Choice A (\(7.00\)) would only occur at the equivalence point of a strong acid-strong base titration.

Adding 0.005 mol HCl converts acetate to acetic acid in both buffers. For Buffer P: [CH3COO^-] drops from 0.010 to 0.005 mol and [CH3COOH] rises to 0.015 mol; pH = 4.74 + log(0.005/0.015) = 4.74 - 0.48 = 4.26 (a change of 0.48 units). For Buffer Q: [CH3COO^-] drops from 1.000 to 0.995 mol and [CH3COOH] rises to 1.005 mol; pH = 4.74 + log(0.995/1.005) ≈ 4.74 - 0.004 = 4.74 (a change of less than 0.01 units). The ratio changes dramatically in Buffer P but barely at all in Buffer Q. This demonstrates that buffer capacity depends on the absolute moles of buffering components, not just their ratio.