1 question, one for each idea where we can. Answer them, then see which ideas to fix.
Question 1 of 1
A buffer solution contains 0.40 mol of CH₃COOH and 0.60 mol of CH₃COO⁻ in 1.0 L. When 0.010 mol of HCl is added, the reading on a pH meter that displays one decimal place does not change. Based on the Henderson-Hasselbalch equation, which statement best explains why?
Answer and reasoning
AThe ratio [CH₃COO⁻]/[CH₃COOH], and so its log, changes only slightlyCorrect The added H₃O⁺ converts a little CH₃COO⁻ into CH₃COOH. Both are present in large amounts, so the ratio [CH₃COO⁻]/[CH₃COOH] and its logarithm change only slightly, and so does pH = pKa + log([CH₃COO⁻]/[CH₃COOH]).
BThe ratio [CH₃COO⁻]/[CH₃COOH] stays the same, as HCl reacts with neither A student who thinks added acid does not react with the buffer components picks this. The added H₃O⁺ reacts with CH₃COO⁻ to form CH₃COOH, so the ratio does decrease, by a small amount.
CThe pKa of CH₃COOH shifts by just enough to offset the added H₃O⁺ A student who thinks an equilibrium constant changes when concentrations change picks this. Ka and pKa depend only on temperature; the pH changes little because the ratio of the two components changes little.
DThe pH of a buffer equals the pKa of its acid, which HCl does not change A student who thinks a buffer's pH is always its pKa picks this. This buffer's pH is above its pKa, because [CH₃COO⁻] > [CH₃COOH]; its pH is set by the pKa and the ratio of the two components.
Working No pH calculation. CH₃COO⁻ + H₃O⁺ → CH₃COOH + H₂O changes the amounts to 0.59 mol CH₃COO⁻ and 0.41 mol CH₃COOH; the ratio [CH₃COO⁻]/[CH₃COOH] falls only slightly (from 1.5 to about 1.44), so log of the ratio, and the pH, change very little. pKa is unchanged.
In preparation: 0 of 1 sections compiled and reviewed. The rest show key terms and common mistakes from our question bank until they are.
8.9.A.1 Henderson-Hasselbalch equation Fix
Henderson-Hasselbalch equation
pH = pKa + log([A⁻]/[HA]). It gives the pH of a buffer from the pKa of the weak acid and the concentration ratio of the conjugate base to the weak acid.
Origin of the Henderson-Hasselbalch equation
The equation is a rearranged, logarithmic form of the Ka expression for the weak acid, Ka = [H₃O⁺][A⁻]/[HA]; it is not a separate law.
Buffer pH and the ratio [A⁻]/[HA]
When [A⁻] = [HA], log(1) = 0 and pH = pKa. When [A⁻] > [HA], pH > pKa; when [A⁻] < [HA], pH < pKa. Each tenfold change in the ratio changes the pH by 1 unit.
Buffer of a weak base and its conjugate acid
For a buffer such as NH₃/NH₄⁺, the Henderson-Hasselbalch equation uses the pKa of the conjugate acid, NH₄⁺, and [A⁻]/[HA] becomes [NH₃]/[NH₄⁺]. At 25°C, pKa = 14.00 − pKb of the base.
Small additions to a buffer
Adding a small amount of strong acid or strong base converts a little of one buffer component into the other. Because both components are present in large amounts, the ratio [A⁻]/[HA] barely changes, so the pH changes much less than it would without the buffer.
Ratio of amounts in the same solution
Because HA and A⁻ are in the same volume of solution, the ratio of their concentrations equals the ratio of their amounts in moles (or of their numbers of particles).
Students often think The Henderson-Hasselbalch equation is pH = pKa + log([HA]/[A⁻]), so a buffer with more conjugate base than acid has a pH below the pKa. In fact No. pH = pKa + log([A⁻]/[HA]): the conjugate base is in the numerator. A buffer with more conjugate base than acid has a pH above the pKa.
Students often think The pH of a buffer equals the pKa of its weak acid, whatever the concentrations of the acid and its conjugate base. In fact No. pH = pKa only when [A⁻] = [HA]. A buffer's pH can be set above or below the pKa by choosing the ratio [A⁻]/[HA]; a tenfold ratio moves it by 1 unit.
6 more questions. Every wrong answer here is a real mistake students make, and you see why it is wrong as soon as you answer.
Question 1 of 6
A buffer solution at 25°C is 0.20 M in NH₃ and 0.10 M in NH₄Cl. The Kb of NH₃ is 1.8 × 10⁻⁵ (pKb = 4.74). What is the pH of the buffer?
Answer and reasoning
A5.04 A student who puts the pKb of NH₃ into the Henderson-Hasselbalch equation in place of the pKa picks this: 4.74 + log 2. The equation needs the pKa of NH₄⁺, 9.26; a buffer that is mostly NH₃ cannot be acidic.
B8.96 A student who puts the acid, NH₄⁺, in the numerator of the ratio picks this: 9.26 + log(0.10/0.20). The base, NH₃, belongs in the numerator, so the excess of NH₃ raises the pH above 9.26.
C9.26 A student who thinks a buffer's pH always equals the pKa of its acid picks this. That is true only when [NH₃] = [NH₄⁺]; here the ratio is 2, which adds log 2 = 0.30.
D9.56Correct The acid member of the pair is NH₄⁺, with pKa = 14.00 − 4.74 = 9.26. pH = 9.26 + log(0.20/0.10) = 9.56; the base, NH₃, is in excess, so the pH is above the pKa.
The titration curve shown was obtained when 25.0 mL of a 0.100 M solution of a weak acid, HA, was titrated with 0.100 M NaOH(aq) at 25°C. Based on the curve, what is the pH of a buffer solution of HA and NaA in which [A⁻]/[HA] = 0.10 at 25°C?
Answer and reasoning
A5.6 A student who thinks a buffer's pH always equals the pKa picks this. pH = pKa only when [A⁻] = [HA]; with [A⁻]/[HA] = 0.10, the log term is −1.0.
B4.6Correct The pKa equals the pH at the half-equivalence point. The equivalence point is at 25.0 mL, so at 12.5 mL the curve gives pKa ≈ 5.6. pH = 5.6 + log(0.10) = 4.6.
C6.6 A student who puts [HA] in the numerator of the ratio picks this: 5.6 + log(10). With less A⁻ than HA, the pH must be below the pKa.
D8.1 A student who reads the pKa at the equivalence point, about pH 9.1, picks this: 9.1 − 1.0. The pKa is the pH at the half-equivalence point, 12.5 mL, which is about 5.6.
Working Equivalence point at 25.0 mL (steep rise); half-equivalence at 12.5 mL, where the curve reads pH ≈ 5.6, so pKa ≈ 5.6. pH = pKa + log([A⁻]/[HA]) = 5.6 + log(0.10) = 5.6 − 1.0 = 4.6. (Equivalence-point pH ≈ 9.1.)
The diagram represents the HA molecules and A⁻ ions in a small volume of a buffer made from the weak acid HA, with pKa = 4.74, and the salt NaA at 25°C. Water molecules and Na⁺ ions are not shown. What is the pH of the buffer?
Answer and reasoning
A4.26 A student who puts HA in the numerator of the ratio picks this: 4.74 + log(2/6). With more A⁻ than HA, the pH must be above the pKa.
B4.62 A student who uses the total acid, HA + A⁻, in the denominator picks this: 4.74 + log(6/8). The ratio compares A⁻ with HA only: 6/2 = 3.0.
C5.22Correct Both species are in the same volume, so [A⁻]/[HA] equals the particle ratio, 6/2 = 3.0. pH = 4.74 + log 3.0 = 5.22, above the pKa because A⁻ is in excess.
D7.74 A student who treats pH − pKa as the ratio itself picks this, adding 3.0 to the pKa. The equation adds the logarithm of the ratio: log 3.0 = 0.48, so pH = 5.22.
Working Count: 2 HA, 6 A⁻. Same volume, so [A⁻]/[HA] = 6/2 = 3.0. pH = 4.74 + log 3.0 = 4.74 + 0.48 = 5.22.
A student needs a buffer with a pH of 5.04 at 25°C and has solutions of CH₃COOH and NaCH₃COO of various concentrations. The pKa of acetic acid is 4.74. Which procedure produces the required buffer?
Answer and reasoning
AMix 50 mL of 0.20 M CH₃COOH with 50 mL of 0.10 M NaCH₃COO A student who puts the acid in the numerator of the ratio picks this, making [CH₃COOH]/[CH₃COO⁻] = 2.0. That buffer has pH = 4.74 + log 0.50 = 4.44, below the pKa.
BMix 100 mL of 0.10 M CH₃COOH with 100 mL of 0.20 M NaCH₃COOCorrect The required ratio is [CH₃COO⁻]/[CH₃COOH] = 105.04 − 4.74 = 100.30 = 2.0. Equal volumes of 0.10 M acid and 0.20 M salt give that ratio, with large amounts of both components.
CMix 100 mL of 0.10 M CH₃COOH with 100 mL of 0.030 M NaCH₃COO A student who takes pH − pKa = 0.30 as the ratio itself picks this. 0.30 is the logarithm of the ratio; the ratio is 100.30 = 2.0.
DAdd water to 0.10 M CH₃COOH until a pH meter reads 5.04 A student who thinks any solution with the right pH is a buffer picks this. The diluted acid has the right pH but contains almost no CH₃COO⁻ and very little CH₃COOH, so it cannot hold its pH when acid or base is added.
Working log([CH₃COO⁻]/[CH₃COOH]) = 5.04 − 4.74 = 0.30, so [CH₃COO⁻]/[CH₃COOH] = 100.30 = 2.0. 100 mL of 0.10 M CH₃COOH + 100 mL of 0.20 M NaCH₃COO gives 0.050 M CH₃COOH and 0.10 M CH₃COO⁻: ratio 2.0, pH 5.04.
The first diagram shown represents the HA molecules and A⁻ ions in a small volume of a buffer solution. A small amount of HCl(aq) is then added, supplying one H₃O⁺ ion to the volume shown. Which of the numbered diagrams best represents the particles in that volume once the mixture has come to equilibrium? Na⁺ ions, Cl⁻ ions and water molecules are not shown.
Answer and reasoning
ADiagram 1Correct The added H₃O⁺ reacts essentially completely with the conjugate base: A⁻ + H₃O⁺ → HA + H₂O. One A⁻ ion is converted to one HA molecule, giving six HA and four A⁻, and no added H₃O⁺ is left. The amount of acid added is exaggerated in the diagram so that the change can be counted; in a real buffer a small addition changes the ratio [A⁻]/[HA], and so the pH, only slightly.
BDiagram 2 A student who thinks added strong acid reacts with the HA of the buffer, using it up and leaving more A⁻, picks this. HA is an acid and does not react with H₃O⁺; the added H₃O⁺ protonates the base A⁻, so HA increases to six and A⁻ falls to four. Four HA and six A⁻ is what adding a little strong base would give.
CDiagram 3 A student who thinks a buffer takes up added acid with no change in its own composition picks this, the same five HA and five A⁻ as before. The H₃O⁺ is removed by reacting with A⁻, which converts one A⁻ into one HA; the pH changes little because the ratio [A⁻]/[HA] changes only modestly, not because nothing changes.
DDiagram 4 A student who thinks a strong acid added to a buffer reacts with neither component picks this, leaving five HA, five A⁻ and the added H₃O⁺ unreacted. A⁻ is a base and reacts essentially completely with the H₃O⁺, so one A⁻ becomes one HA and no added H₃O⁺ remains.
Working Before the addition there are five HA and five A⁻. The added H₃O⁺ reacts essentially completely with the base A⁻: A⁻ + H₃O⁺ → HA + H₂O. One A⁻ is converted to one HA, so the volume then holds six HA and four A⁻ with no added H₃O⁺ left over.
A buffer solution at 25°C is 0.50 M in a weak acid, HA, and 0.25 M in its salt, NaA. The pKa of HA is 5.20. A 10.0 mL sample of the buffer is diluted with distilled water to a total volume of 100.0 mL. What is the pH of the diluted solution?
Answer and reasoning
A5.20 A student who thinks the pH of a buffer equals the pKa of its weak acid whatever the concentrations picks 5.20. The pH equals the pKa only when [A⁻] = [HA]; here [A⁻]/[HA] = 0.50, so the pH is 0.30 unit below the pKa.
B5.90 A student who thinks a tenfold dilution raises the pH of any acidic solution by one unit finds the buffer's pH, 4.90, and adds 1.00. That rule holds for a strong acid; in a buffer the ratio [A⁻]/[HA] is unchanged by dilution, so the pH stays at 4.90.
C5.40 A student who treats the buffer like a solution of the weak acid alone adds half a pH unit for the tenfold dilution, 4.90 + 0.50. With A⁻ present in a large amount, [H₃O⁺] = Ka[HA]/[A⁻], which depends only on the ratio, so the pH stays at 4.90.
D4.90Correct Dilution lowers [HA] and [A⁻] by the same factor, so [A⁻]/[HA] stays at 0.50 and pH = 5.20 + log 0.50 = 4.90, the same as before the water was added.
Working Dilution lowers both concentrations by the same factor of 10: [HA] = 0.050 M and [A⁻] = 0.025 M, so the ratio [A⁻]/[HA] is still 0.25/0.50 = 0.50. pH = pKa + log([A⁻]/[HA]) = 5.20 + log 0.50 = 5.20 − 0.30 = 4.90, the same as before dilution. (Both concentrations remain far greater than [H₃O⁺] = 1.3 × 10⁻⁵ M, so the equation still applies.)
Compiled from the AP Chemistry Course and Exam Description (effective Fall 2024) and our question bank · Specialist review in progress. How these pages are made · Free, no account
