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AP Unit 8 Solutions

Worked solutions to all 15 Unit 8 problems. These answers and reasons were added for the AP edition. Check the setup, units, assumptions and explanation alongside the final result.

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Problem 1

At a particular temperature, Kw = 4.00 × 10−14. Calculate [H3O+], [OH−], pH, and pOH in pure water at that temperature. A student calls this water acidic because its pH is below 7. Evaluate the claim and state the correct pH + pOH relation.

  1. Neutral pure water has equal hydronium and hydroxide concentrations. Let both be x: x2 = Kw, so x = 2.00 × 10−7 M.
  2. pH = pOH = −log(2.00 × 10−7) = 6.699.
  3. pKw = −log(4.00 × 10−14) = 13.398, and pH + pOH = pKw.
  4. This water is neutral because the two ion concentrations are equal. The neutral pH of 7.00 and sum of 14.00 apply at 25 °C when Kw = 1.0 × 10−14.

Final answer

[H3O+] = [OH−] = 2.00 × 10−7 M; pH = pOH = 6.699; sum = 13.398; the water is neutral.

What earns credit
  • Use equal neutral-ion concentrations and x² = Kw to obtain 2.00 × 10−7 M.
  • Calculate pH = pOH and pKw; identify neutrality by equal ion concentrations rather than the number 7.

Watch out: Neutrality always means pH 7 and pH+pOH 14, regardless of temperature.

Return to problem 1 → · Review Topic 8.1

Problem 2

At 25 °C, calculate pH and pOH for (a) 0.00300 M HNO3 and (b) a fully dissolved 0.00300 M Ba(OH)2 solution. Show the major solute ions and their concentrations for each. Explain why “strong” does not mean “concentrated.”

  1. HNO3 ionizes essentially completely: [H3O+] ≈ [NO3−] = 0.00300 M. pH = 2.523 and pOH = 11.477.
  2. Ba(OH)2 dissociates to Ba2+ + 2OH−: [Ba2+] = 0.00300 M and [OH−] = 0.00600 M. pOH = 2.222 and pH = 11.778.
  3. Strong describes essentially complete ionization/dissociation of the dissolved solute. Concentrated describes how much solute is present per volume; a strong acid can be dilute. Water-derived ions are negligible compared with these millimolar concentrations.

Final answer

(a) pH 2.523, pOH 11.477; (b) pH 11.778, pOH 2.222.

What earns credit
  • Identify complete monoprotic ionization and the 2:1 OH−/Ba(OH)2 stoichiometry.
  • Calculate both pH/pOH pairs at 25 °C and specify major spectator-ion concentrations.
  • Distinguish extent of ionization from analytical concentration.

Watch out: One formula unit of every hydroxide produces one hydroxide ion.

Return to problem 2 → · Review Topic 8.2

Problem 3

A 0.100 M solution of a monoprotic weak acid HA has Ka = 1.00 × 10−5 at 25 °C. Determine pH, percent ionization, the major acid-related species, [OH−], and Kb of A−. Show and check the small-x approximation. Predict qualitatively what dilution does to percent ionization and hydronium concentration.

  1. For HA + H2O ⇌ H3O+ + A−, let x ≈ [H3O+] = [A−]. Ka = x2/(0.100 − x).
  2. Assuming x ≪ 0.100 gives x = √(1.00 × 10−5 × 0.100) = 1.00 × 10−3 M. x/C = 1.00%, so the approximation is acceptable; the exact root is 9.95 × 10−4 M.
  3. Using the exact root, pH = 3.002, percent ionization = 0.995%, [HA] ≈ 0.0990 M, [A−] ≈ 9.95 × 10−4 M, and [OH−] ≈ 1.01 × 10−11 M.
  4. Kb(A−) = Kw/Ka = 1.00 × 10−9. Dilution increases the fraction ionized, while decreasing hydronium concentration in this ordinary weak-acid regime: more fraction of a smaller total is not necessarily more ions per liter.

Final answer

pH ≈ 3.00; ionization ≈ 1.00%; mostly HA. Kb = 1.00 × 10−9; dilution raises the ionized fraction but lowers hydronium concentration.

What earns credit
  • Write Ka = x²/(C−x) and justify the small-change check.
  • Give hydronium, percent ionization and remaining HA consistently; apply Kw for hydroxide and the conjugate Kb.
  • Explain opposite dilution effects on fractional ionization and ion concentration without changing Ka.

Watch out: Dilution makes Ka increase or necessarily increases hydronium concentration.

Return to problem 3 → · Review Topic 8.3

Problem 4

A 0.0200 M weak base B has Kb = 2.00 × 10−5 at 25 °C. Estimate pOH, pH, percent protonation, and Ka of its conjugate acid HB+. Write the equilibrium equation and show that the approximation is reasonable. Describe the particle population.

  1. B + H2O ⇌ HB+ + OH−. Kb = x2/(0.0200 − x), where x is the concentration of each formed ion.
  2. x ≈ √(2.00 × 10−5 × 0.0200) = 6.32 × 10−4 M. The fractional change is 3.16%, below a 5% classroom check.
  3. Approximate pOH = 3.199, pH = 10.801, and percent protonation = 3.16%. Most B molecules remain B; much smaller equal amounts of HB+ and OH− form.
  4. Ka(HB+) = Kw/Kb = 5.00 × 10−10. These Ka and Kb belong to the same conjugate pair, not to unrelated acids and bases.

Final answer

Approximately pOH = 3.20, pH = 10.80, protonation = 3.16%, and Ka = 5.00 × 10−10.

What earns credit
  • Write the weak-base proton-transfer equation and identify x as hydroxide.
  • Compute the checked estimate, pOH, pH and protonated fraction.
  • Identify mostly B and use the paired KaKb relation for HB+.

Watch out: A weak base makes [OH−] equal to its initial concentration.

Return to problem 4 → · Review Topic 8.3

Problem 5

At 25 °C, 25.0 mL of 0.120 M HCl is mixed with 35.0 mL of 0.100 M NaOH. Determine the excess reagent, [OH−], pH, and the major spectator-ion concentrations. Assume additive volumes. Explain why subtracting molarities would be wrong.

  1. Moles acid = 0.0250 × 0.120 = 0.00300 mol; moles OH− = 0.0350 × 0.100 = 0.00350 mol. Hydronium and hydroxide react essentially quantitatively 1:1.
  2. OH− excess = 0.000500 mol. Total volume = 0.0600 L, so [OH−] = 0.00833 M.
  3. pOH = 2.079, so pH = 11.921. [Na+] = 0.00350/0.0600 = 0.0583 M; [Cl−] = 0.00300/0.0600 = 0.0500 M. Their charge difference balances excess OH−.
  4. The starting volumes differ, so the reagent amount comparison must use moles. Only after reaction can the remaining moles be divided by the common final volume.

Final answer

Excess OH−; [OH−] = 0.00833 M, pH = 11.92; [Na+] 0.0583 M, [Cl−] 0.0500 M.

What earns credit
  • Compare acid/base moles with a 1:1 neutralization equation.
  • Divide excess hydroxide by combined volume before calculating pH.
  • Determine spectator concentrations and explain why unequal stock volumes forbid direct molarity subtraction.

Watch out: Subtract initial concentrations instead of reacting initial moles.

Return to problem 5 → · Review Topic 8.4

Problem 6

Three separate 25.0 mL samples of 0.100 M HA, Ka = 1.00 × 10−5, receive 0.100 M NaOH at 25 °C: sample I gets 10.0 mL, II gets 25.0 mL, III gets 30.0 mL. Identify major acid-related species and select the pH method for each. Calculate pH for II and III; do not compute a buffer pH change for I.

  1. Each begins with 2.50 mmol HA. HA + OH− → A− + H2O occurs essentially quantitatively. I receives 1.00 mmol OH−, leaving 1.50 mmol HA and producing 1.00 mmol A−: a buffer. Its current pH can be found from the conjugate-pair ratio.
  2. II receives 2.50 mmol OH−, converting essentially all initial HA to A−. [A−] formal = 2.50 mmol/50.0 mL = 0.0500 M. A− + H2O ⇌ HA + OH− has Kb = Kw/Ka = 1.00 × 10−9.
  3. For II, [OH−] ≈ √(1.00 × 10−9 × 0.0500) = 7.07 × 10−6 M; pH ≈ 8.85. The conjugate base makes equivalence basic.
  4. III has 0.500 mmol excess OH− in 55.0 mL, so [OH−] = 0.00909 M and pH = 11.96. Excess strong base dominates, not Henderson–Hasselbalch.

Final answer

I: HA/A− buffer. II: A−-dominated weak-base equilibrium, pH ≈ 8.85. III: excess strong base, pH ≈ 11.96.

What earns credit
  • Identify the three remaining-species regimes from moles: buffer, conjugate-base equivalence, excess strong base.
  • Use Kb = Kw/Ka and the diluted A− concentration for the equivalence pH.
  • Use excess OH− and total volume for the beyond-equivalence pH.

Watch out: At equivalence every acid–base mixture has pH 7.

Return to problem 6 → · Review Topic 8.4

Problem 7

A weak base B has Kb = 1.00 × 10−5. Explain the dominant species and pH method when a 20.0 mL 0.100 M B sample receives 0.100 M HCl volumes of 5.0, 20.0, or 25.0 mL at 25 °C. Then consider a separate mixture HA + B ⇌ A− + HB+. Ka(HA) = 1.00 × 10−4 and Ka(HB+) = 1.00 × 10−8. Is quantitative neutralization always justified for arbitrary weak–weak mixtures? Use these constants to support the reasoning for this example.

  1. B starts at 2.00 mmol. With 0.500 mmol HCl it leaves 1.50 mmol B and makes 0.500 mmol HB+: a buffer. With 2.00 mmol HCl it yields mostly HB+: use Ka(HB+) = Kw/Kb = 1.00 × 10−9 to describe the acidic equilibrium.
  2. With 2.50 mmol HCl, 0.500 mmol hydronium remains in 45.0 mL: excess strong acid determines pH; [H3O+] ≈ 0.0111 M.
  3. For HA + B ⇌ A− + HB+, adding HA dissociation to the reverse of HB+ dissociation gives K = Ka(HA)/Ka(HB+) = 104. The example strongly favors products.
  4. Other weak–weak pairs can have K near one or below one, so a quantitative stoichiometric completion assumption is not universal. Treat their mixture as a proton-transfer equilibrium based on relative acid strengths.

Final answer

5.0 mL: buffer; 20.0 mL: acidic HB+ equilibrium; 25.0 mL: excess strong acid. The weak–weak example has K = 104, but arbitrary weak–weak pairs need equilibrium reasoning.

What earns credit
  • Identify B/HB+ buffer, HB+ equivalence and excess acid from reagent amounts.
  • Relate HB+ acidity to B basicity.
  • Obtain the weak–weak proton-transfer K by combining acid equilibria and avoid a universal completion claim.

Watch out: All weak acid–weak base mixtures neutralize quantitatively regardless of their strengths.

Return to problem 7 → · Review Topic 8.4

Problem 8

At 25 °C, a 25.00 mL monoprotic weak-acid sample is titrated with 0.1000 M NaOH. Supplied data are: added volume 0.00, 10.00, 19.50, 20.00, 20.50, 25.00 mL; pH 2.92, 4.80, 6.39, 8.72, 11.04, 12.00. The equivalence volume from the steep region is 20.00 mL.

  1. Determine the acid’s initial concentration and pKa. Identify major acid-related species at 10.00 and 20.00 mL, write the conjugate base’s reaction with water, and explain why the equivalence pH exceeds 7.
  2. Predict the bias in calculated acid concentration if a delivery-tip air bubble fills during titration, so recorded volume exceeds delivered volume.
  3. Two separate titrations at 25 °C are stopped exactly at their stoichiometric equivalence points: NH3(aq) with HCl(aq), and HCl(aq) with NaOH(aq). For each, identify the major solute ions, write the relevant proton-transfer reaction with water or explain why there is no appreciable one, and predict whether pH is below, equal to, or above 7. Explain why the same definition of equivalence can give different pH values.

Assume dilute aqueous solutions, complete strong-acid/strong-base dissociation, no excess titrant, and no competing reactions. Chloride and sodium ions have negligible acid–base effects in this model.

  1. At equivalence for the monoprotic weak acid, initial acid moles = 0.1000 mol L−1 × 0.02000 L = 0.002000 mol. Initial concentration = 0.002000/0.02500 = 0.08000 M.
  2. 10.00 mL is half the equivalence volume. Comparable HA and A− amounts remain, so pH ≈ pKa = 4.80.
  3. At 20.00 mL the major acid-related species is A−. A− + H2O ⇌ HA + OH− produces hydroxide, giving a basic equivalence solution. pH 7 is not universal at equivalence.
  4. A delivery-tip bubble that fills uses some indicated buret volume without delivering it to the flask. Using that inflated volume overestimates acid moles and initial acid concentration.
  5. For NH3 titrated with HCl, the essentially quantitative reaction NH3 + H3O+ → NH4+ + H2O leaves NH4+ and Cl− as major solute ions at equivalence. NH4+ + H2O ⇌ NH3 + H3O+ makes the solution acidic, with pH below 7 at 25 °C. There is no excess strong acid; the hydronium comes from the conjugate acid’s equilibrium.
  6. For HCl titrated with NaOH, H3O+ + OH− → 2H2O leaves Na+ and Cl−. Neither undergoes appreciable proton transfer in the stated model. Water autoionization gives equal hydronium and hydroxide concentrations: neutral pH = 7.00 at 25 °C.
  7. Equivalence fixes reacting amounts by stoichiometry. The pH then depends on the remaining species and their reactions with water. At other temperatures, a neutral solution still has equal [H3O+] and [OH−], but its pH need not be 7.

Final answer

Initial weak-acid concentration = 0.08000 M; pKa ≈ 4.80; half-equivalence has HA/A−, equivalence mostly A−. The bubble error biases concentration high. NH3/HCl equivalence is acidic from NH4+; HCl/NaOH equivalence is neutral at pH 7.00 at 25 °C.

What earns credit
  • Use monoprotic equivalence stoichiometry to calculate original concentration and read pKa at half-equivalence.
  • Identify A− at weak-acid equivalence and write A− + H2O ⇌ HA + OH− to explain basic pH.
  • Trace the filling delivery-tip bubble to overreported delivery and a high concentration result.
  • Identify NH4+/Cl− and write ammonium proton transfer with water to explain acidic weak-base equivalence without excess strong acid.
  • Identify Na+/Cl−, negligible ion hydrolysis and equal hydronium/hydroxide for neutral strong/strong equivalence; qualify pH 7 by 25 °C.

Watch out: The equivalence pH is always 7, and every buret volume difference is volume delivered to the flask.

Return to problem 8 → · Review Topic 8.5

Problem 9

A hypothetical diprotic acid H2A is titrated with strong base. Its resolved curve has equivalence points at 15.0 and 30.0 mL; pH values at 7.50 and 22.5 mL are 3.00 and 7.00. Infer the number of distinguishable acidic protons, pKa1 and pKa2, and the major acid-related species at 0, 15.0, 22.5, and 30.0 mL. Explain a limitation of inferring proton count from an unresolved titration curve. No species-concentration calculations are required.

  1. Two resolved equivalence regions with equal volume increments support two sequentially titratable protons. The second equivalence requires twice the base amount of the first.
  2. At the halfway point of the first step, pH ≈ pKa1 = 3.00. Halfway between the equivalence points, pH ≈ pKa2 = 7.00.
  3. Initially H2A predominates; near first equivalence HA− predominates; at 22.5 mL HA− and A2− are present comparably; near second equivalence A2− predominates.
  4. If successive Ka values are too close or a step is too weak to resolve in the measured range, the curve may not show distinct jumps for every acidic proton. Absence of a visible jump is not universal proof of absence of a proton.

Final answer

Two resolved protons; pKa1 ≈ 3.00, pKa2 ≈ 7.00; species progress H2A → HA− → HA−/A2− → A2−.

What earns credit
  • Infer two resolved sequential proton transfers and both half-step pKa values.
  • Identify dominant forms in the four specified regions.
  • State an overlap/resolution limitation without performing polyprotic concentration calculations.

Watch out: Every polyprotic acid must display a visibly separate jump for every proton.

Return to problem 9 → · Review Topic 8.5

Problem 10

  1. Explain why CH3COOH is more acidic than CH3CH2OH and why ClCH2COOH is more acidic than CH3COOH. Identify the transferred proton and compare the resulting conjugate-base structures.
  2. Explain how NH3 acts as a weak base. Identify its proton-accepting site and draw or describe the product.
  3. HNO3 is a strong acid in dilute water. One valid Lewis contributor is H–O–N+(=O)–O−. Identify its transferred proton, draw one valid contributor of its conjugate base, and describe how the ion’s structure stabilizes negative charge and makes it a very weak base. The supplied strong-acid classification is not something you must derive from resonance alone.
  4. NaOH and Ca(OH)2 are ionic hydroxides. Identify the proton-accepting ion released into water, its accepting atom and conjugate acid. Explain why the dissolved portion gives strong-base behavior, and why this does not mean every added piece of Ca(OH)2 must dissolve. Contrast that ion with ammonia’s molecular accepting site.

Use comparable dilute aqueous conditions. Relate strength to structure and proton transfer; hydrogen count alone is not evidence.

  1. The carboxylic acids transfer the carboxyl O–H proton; ethanol transfers its hydroxyl O–H proton. The products are CH3COO−, ClCH2COO− and CH3CH2O−. Carboxylate resonance delocalizes negative charge across two oxygens, unlike ethoxide. This stabilizes acetate and favors proton loss from acetic acid relative to ethanol under comparable conditions.
  2. Electronegative Cl additionally withdraws electron density through bonds, stabilizing chloroacetate relative to its acid and increasing acidity compared with acetic acid.
  3. NH3 accepts a proton using a lone pair on N, forming NH4+. NH3 + H2O ⇌ NH4+ + OH− is incomplete under ordinary dilute aqueous conditions, so ammonia is a weak base.
  4. HNO3 transfers its O–H proton. One nitrate contributor is O−–N+(=O)–O−, with overall charge −1. There are three equivalent contributors placing the N=O bond on different oxygens; the actual ion is a resonance hybrid, not an ion alternating between drawings. Negative charge is distributed over electronegative oxygen atoms, stabilizing NO3−. As the conjugate base of a strong acid it has very little tendency to take a proton from water. Resonance is part of this structural explanation, not by itself a universal test for a strong acid.
  5. NaOH gives Na+ and OH−; dissolved Ca(OH)2 gives Ca2+ and two OH− per formula unit. OH− accepts a proton at oxygen to form H2O, its very weak conjugate acid in water. The dissolved ionic hydroxide dissociates essentially completely, unlike ammonia’s incomplete production of OH− by proton transfer.
  6. Calcium hydroxide’s limited solubility can leave solid behind. Base strength of its dissolved portion and total amount dissolved are different questions. Sodium and calcium ions are spectators in the stated acid–base model; ammonia instead accepts directly at a nitrogen lone pair.

Final answer

Carboxyl O–H proton loss gives resonance-stabilized carboxylates; Cl adds inductive stabilization. NH3 accepts at its N lone pair. HNO3 gives resonance-stabilized NO3−, a very weak base. Dissolved NaOH/Ca(OH)2 provide OH−, which accepts at O to form water; complete dissociation of the dissolved portion does not imply unlimited solubility.

What earns credit
  • Identify acid O–H sites, ethoxide and carboxylate products; use resonance and chlorine’s inductive effect causally.
  • Identify ammonia’s N lone pair, NH4+ product and incomplete proton transfer with water.
  • Identify nitric acid’s O–H proton, a charge-correct nitrate contributor, equivalent resonance contributors and very weak conjugate-base behavior.
  • Identify hydroxide’s O accepting site and H2O conjugate acid, with correct NaOH/Ca(OH)2 ion stoichiometry.
  • Distinguish complete dissociation of dissolved hydroxide from limited total solubility; do not equate strong with concentrated.

Watch out: Resonance alone proves every acid is strong, or strong-base behavior means unlimited solid solubility.

Return to problem 10 → · Review Topic 8.6

Problem 11

A weak acid indicator has pKa = 6.50. Its protonated HIn form is yellow and deprotonated In− form is blue. Predict the predominant form at pH 5.50 and 7.50 and the approximate ratio [In−]/[HIn] at each. Would this indicator be a good choice for a titration whose sharply rising equivalence region runs from pH 8.0 to 10.0? Explain using the curve region, not merely whether a color change exists.

  1. For a conjugate pair, [In−]/[HIn] = 10^(pH − pKa). At 5.50 the ratio is 0.10, so yellow HIn predominates. At 7.50 it is 10, so blue In− predominates.
  2. The visible transition occurs around pKa as both forms become appreciable, roughly pKa ± 1 for a conventional two-color indicator.
  3. Its transition is largely around pH 5.5–7.5, before the stated steep equivalence region. It would tend to signal too early and is a poor match.
  4. Choose an indicator whose transition falls within the steep region and whose pKa is close to the equivalence pH. Color changes show protonation state, not a universal pH endpoint.

Final answer

pH 5.50: yellow HIn, ratio 0.10. pH 7.50: blue In−, ratio 10. This indicator is poorly matched to the pH 8.0–10.0 equivalence region.

What earns credit
  • Calculate each indicator-form ratio with the correct sign and identify the corresponding color.
  • Compare the transition interval with the actual steep equivalence region.
  • Reject a mismatched endpoint using titration evidence.

Watch out: Any indicator is suitable because all indicators change color somewhere.

Return to problem 11 → · Review Topic 8.7

Problem 12

Three equal-volume solutions contain (I) 0.100 M HA only, (II) 0.100 M HA and 0.100 M A−, or (III) 0.100 M HCl and 0.100 M NaCl. HA is a weak acid. Which is an effective buffer near pKa(HA)? Write the reactions that consume a small added quantity of strong acid and strong base in that solution. Explain why its pH changes slightly rather than remaining perfectly fixed, and why the others are less suitable.

  1. II has appreciable amounts of a weak conjugate pair. Added hydronium reacts with A−: A− + H3O+ → HA + H2O. Added hydroxide reacts with HA: HA + OH− → A− + H2O.
  2. These reactions convert one buffer form into the other, limiting the increase in free hydronium or hydroxide. The pair ratio changes somewhat, so pH changes somewhat.
  3. I contains mostly HA and little A− from its own ionization, so it cannot neutralize added strong acid as effectively as II.
  4. III uses a strong acid and its extremely weak conjugate base; Cl− does not appreciably consume added acid, so this is not an effective weak-pair buffer.

Final answer

II is the effective buffer; A− consumes added acid and HA consumes added base. Capacity is finite and pH is stabilized, not frozen.

What earns credit
  • Select the weak conjugate pair with appreciable amounts of both forms.
  • Write the correct acid-consuming and base-consuming equations.
  • Explain finite ratio change and diagnose why weak-acid-only and HCl/Cl− solutions are poor alternatives.

Watch out: Any acid and its conjugate base, including HCl/Cl−, makes an effective buffer.

Return to problem 12 → · Review Topic 8.8

Problem 13

A buffer has pKa(HA) = 4.80, [HA] = 0.200 M, and [A−] = 0.0500 M. Calculate its current pH and identify the more abundant form. To prepare a buffer at pH 5.40 using the same pair, what concentration ratio [A−]/[HA] is needed? State the condition under which using a mole ratio would give the same result. Do not calculate a pH change after an acid/base addition.

  1. pH = pKa + log([A−]/[HA]) = 4.80 + log(0.0500/0.200) = 4.20, two decimal places. HA is four times as abundant as A−, consistent with pH below pKa.
  2. For target pH 5.40, log ratio = 5.40 − 4.80 = 0.60; ratio = 100.60 ≈ 4.0.
  3. Both species occupy the same final solution volume, so ([nA−]/V)/([nHA]/V) = nA−/nHA. If mixing or neutralization occurs first, use the resulting pair amounts, not unchanged stock concentrations.
  4. The approximation requires an appreciable conjugate pair and small equilibrium changes relative to the prepared component amounts; use activities for greater rigor.

Final answer

Current pH = 4.20; HA predominates. Target ratio [A−]/[HA] ≈ 4.0; a mole ratio works for the two components in their common final volume.

What earns credit
  • Use the base/acid ratio in Henderson–Hasselbalch and report current pH consistently.
  • Obtain the requested target ratio and verify which form predominates.
  • Explain common-volume cancellation and use composition after preparation reactions.

Watch out: Invert the base/acid ratio or use unrelated stock concentrations in the ratio.

Return to problem 13 → · Review Topic 8.9

Problem 14

Equal 100 mL buffers A and B have the same HA/A− pair. A contains 0.100 M of each component; B contains 0.0100 M of each. Compare their initial pH and ability to resist the same added amount of acid. Buffer C instead contains 0.150 M HA and 0.0500 M A−. Compare C’s ability to consume added acid versus added base. If B was accidentally diluted further before an investigation, explain the expected bias in its apparent capacity while the ratio remains unchanged.

  1. A and B have the same ratio 1:1, so both have pH ≈ pKa. A has ten times as many moles of each component in the equal volumes and can neutralize much more added acid/base before a form is depleted.
  2. C has more HA than A−. HA consumes added base, so C has greater reserve for added base than added acid. Its acid-consuming reserve is the smaller A− amount.
  3. Accidental uniform dilution preserves the ratio approximately but lowers component concentration. For the same test volume and added moles it lowers the apparent capacity; similar starting pH does not reveal the missing reservoir.
  4. A comparison must specify volume and added amount: total moles matter to a whole sample, while concentration describes capacity per volume.

Final answer

A and B start near the same pH; A has larger capacity. C better resists added base. Uniform accidental dilution lowers measured capacity per test volume without much initial-pH change.

What earns credit
  • Separate ratio-controlled pH from amount-controlled capacity at equal volume.
  • Identify HA as the reserve for added base and A− as the reserve for added acid.
  • Explain why uniform unintended dilution lowers measured capacity while preserving approximate pH.

Watch out: Buffers with the same pH must have the same capacity.

Return to problem 14 → · Review Topic 8.10

Problem 15

  1. Compare adding acid to saturated solutions containing M(OH)2(s), MCO3(s), and MX(s), where X− is the extremely weak conjugate base of a strong acid. Write a linked proton-transfer equation for the hydroxide or carbonate, explain the dissolution response, and state whether Ksp changes at fixed temperature.
  2. A separate hypothetical sparingly soluble salt JHY(s) dissolves as JHY(s) ⇌ JH+(aq) + Y−(aq). JH+ is a weak acid; its neutral conjugate base J remains dissolved. Y− is acid–base inactive. Write a proton-transfer reaction for JH+ with water or hydroxide. Predict the effect of raising pH on free JH+, the immediate dissolution quotient, and saturated total solubility. Distinguish free JH+ from total J-containing dissolved material.
  3. For this JHY model, predict how lowering pH instead affects the JH+/J distribution and saturated total solubility. State one condition under which the simplified prediction would need to be reconsidered.

Assume constant temperature, negligible dilution, a macroscopic amount of the original solid present, and no common-ion addition, complexes, competing precipitates or other acid–base reactions over the specified pH range. These are supplied models, not chemical-handling instructions. No solubility-versus-pH calculation is required.

  1. For the hydroxide, OH− + H3O+ → 2H2O consumes a dissolution product. Lower OH− lowers Qsp, allowing additional M(OH)2 to dissolve.
  2. For the carbonate, CO32− + H3O+ ⇌ HCO3− + H2O protonates the basic ion, removing free carbonate from the dissolution quotient. Additional solid can dissolve. Further protonation can matter depending on conditions.
  3. X− is not appreciably protonated in the stated MX model, so acid does not give this protonation-driven enhancement. Ksp for each specified solid is unchanged at fixed temperature.
  4. For the acidic-ion model, JH+ + H2O ⇌ J + H3O+, or equivalently JH+ + OH− → J + H2O. Raising pH favors conversion of free JH+ to J.
  5. Since Qsp = [JH+][Y−], removal of free JH+ first lowers Qsp below Ksp. The later net dissolution supplies more ions; some JH+ then deprotonates. Saturated total J-containing material, [JH+] + [J], increases. Ksp uses free JH+, not this total, and stays fixed.
  6. Lowering pH favors JH+ relative to J. At the same initial dissolved totals, protonation raises the free acidic-ion share and Qsp; the later response can precipitate JHY until saturation is restored. The saturated total solubility is lower within this model.
  7. If neutral J precipitated, a different solid formed, complexation became significant, or all original solid dissolved, the stated saturated single-solid model would need reconsideration. The result is conditional, not a universal trend for all salts.

Final answer

Acid increases hydroxide/carbonate solubility by consuming a free dissolution product; inactive X− lacks that response. For acidic JH+, raising pH consumes the free ion and increases total JHY solubility; lowering pH favors JH+ and lowers total solubility in the stated model. Ksp stays fixed and uses free ions.

What earns credit
  • Write hydroxide neutralization or carbonate protonation and connect removal of a dissolved product to extra dissolution.
  • Distinguish inactive strong-acid conjugate X− and keep fixed-temperature Ksp unchanged.
  • Write the acidic JH+ proton-transfer equation and explain its consumption when pH rises.
  • Use Qsp = [JH+][Y−] to connect free-ion removal to increased saturated total [JH+] + [J], not to a changed Ksp.
  • Explain the opposite lower-pH distribution/solubility trend and identify a stated model limit or competing phase.

Watch out: A pH effect on total solubility changes Ksp, or acid and base perturbations affect every constituent ion in the same direction.

Return to problem 15 → · Review Topic 8.11

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