8.5 Acid-Base Titrations

1. AAt the equivalence point itself
   A student who thinks HA and A⁻ are equal at the equivalence point picks this. At the equivalence point almost all of the HA has been converted, so the diagram would show A⁻ ions and almost no HA.
2. BAt the half-equivalence point Correct
   Six acid units were present at the start; three have been converted to A⁻ by the added OH⁻ and three remain as HA. Half of the acid has been neutralized, which is the half-equivalence point, where [HA] = [A⁻].
3. CBefore any NaOH has been added
   A student who thinks a weak acid at equilibrium is about half ionized picks this. Before base is added, a weak acid solution is mostly HA with very little A⁻.
4. DAt the point where pH = 7.00
   A student who thinks equal amounts of an acid and its conjugate base give a neutral solution picks this. With [HA] = [A⁻], pH = pK_a of HA, which for a typical weak acid is well below 7.

Working No calculation. Equal numbers of HA and A⁻ (3 and 3): half of the HA originally present (6 units) has been converted to A⁻ by OH⁻. This is the half-equivalence point.

CED 8.5.A.1 · Read this in Fix

1. AMoles of OH⁻ added equal the moles of HA originally present Correct
   The equivalence point is reached when the moles of titrant added equal the moles of monoprotic analyte originally present in the sample, so M(NaOH) × V(NaOH) gives the moles of HA, and dividing by the sample volume gives its concentration.
2. BMoles of HA remaining equal the moles of A⁻ that have formed
   A student who thinks the amounts of HA and A⁻ are equal at the equivalence point picks this. HA and A⁻ are equal at the half-equivalence point; at the equivalence point essentially all of the HA has been converted to A⁻.
3. CConcentration of H₃O⁺ equals the concentration of OH⁻ present
   A student who thinks the solution is neutral at the equivalence point of every titration picks this. At the equivalence point of a weak acid titration the A⁻ present reacts with water, so [OH⁻] is greater than [H₃O⁺], and this comparison says nothing about how much HA the sample contained.
4. DConcentration of the NaOH equals the concentration of the acid
   A student who thinks the analyte and titrant solutions have equal concentrations at the equivalence point picks this. It is the moles that are equal; the concentrations are equal only if the two volumes happen to be equal.

Working HA(aq) + OH⁻(aq) → A⁻(aq) + H₂O(l) in a 1:1 mole ratio. At the equivalence point, moles of OH⁻ added = moles of HA in the sample, so M(NaOH) × V(NaOH at equivalence) = M(HA) × V(HA sample), and M(HA) = M(NaOH) × V(NaOH)/V(HA sample). The relationship holds for weak and strong acids alike.

CED 8.5.A.2 · Read this in Fix

1. A4.5 Correct
   The equivalence point is at 25.0 mL, so the half-equivalence point is at 12.5 mL, where the curve shows pH 4.5. There [HA] = [A⁻], so pH = pK_a = 4.5.
2. B2.8
   A student who thinks pK_a is the pH of the acid before any base is added reads the starting pH and picks this. That pH also depends on the concentration of HA; pK_a is the pH at the half-equivalence point.
3. C8.6
   A student who thinks pK_a is the pH at the equivalence point picks this. That pH is set by A⁻ reacting with water; pK_a is read at half the equivalence volume.
4. D9.5
   A student who thinks the half-equivalence pH gives pK_b of A⁻ calculates pK_a = 14 − 4.5 and picks this. At the half-equivalence point pH = pK_a of the acid HA directly.

Working Equivalence point at 25.0 mL; half-equivalence at 12.5 mL, where pH = 4.5 from the curve. At the half-equivalence point [HA] = [A⁻], so pK_a = pH = 4.5.

CED 8.5.A.3 · Read this in Fix

1. A5.13 Correct
   At the equivalence point 4.0 mmol of NH₄⁺ are in 40.0 mL: [NH₄⁺] = 0.10 M. K_a = K_w/K_b = 5.6 × 10⁻¹⁰, so [H₃O⁺] = √(5.6 × 10⁻¹⁰ × 0.10) = 7.5 × 10⁻⁶ M and pH = 5.13; the conjugate acid NH₄⁺ makes the solution acidic.
2. B2.87
   A student who uses K_b of NH₃ as if it were K_a of NH₄⁺ picks this: √(1.8 × 10⁻⁵ × 0.10). K_a of NH₄⁺ = K_w/K_b = 5.6 × 10⁻¹⁰, giving [H₃O⁺] = 7.5 × 10⁻⁶ M.
3. C4.98
   A student who takes [NH₄⁺] at the equivalence point to be 0.20 M, the original concentration, picks this. Adding 20.0 mL of HCl doubles the volume, so [NH₄⁺] = 0.10 M.
4. D7.00
   A student who thinks every equivalence point is at pH 7 picks this. At this equivalence point the solution contains NH₄⁺, which donates protons to water, so the pH is below 7.

Working n(NH₃) = 4.0 mmol, so 20.0 mL of HCl is needed; total volume 40.0 mL. [NH₄⁺] = 4.0 mmol/40.0 mL = 0.10 M. K_a(NH₄⁺) = 1.0 × 10⁻¹⁴/1.8 × 10⁻⁵ = 5.6 × 10⁻¹⁰. NH₄⁺ + H₂O ⇌ NH₃ + H₃O⁺: [H₃O⁺] ≈ √(5.6 × 10⁻¹⁰ × 0.10) = 7.5 × 10⁻⁶ M; pH = 5.13.

CED 8.5.A.4 · Read this in Fix

1. A1 (monoprotic)
   A student who counts only the large rise near 40 mL as an equivalence point picks this. The smaller rise at 20 mL is also an equivalence point.
2. B3 (triprotic)
   A student who counts the flat regions, including the one after 40 mL, picks this. The flat region after the last equivalence point shows excess NaOH and corresponds to no proton.
3. C2 (diprotic) Correct
   The curve has two steep rises, at 20 mL and at 40 mL, equally spaced. Each equivalence point marks the removal of one proton, so each molecule has two acidic protons.
4. D4 (tetraprotic)
   A student who counts every bend in the curve, including the middle of each flat region, picks this. Only the two steep rises are equivalence points.

Working Count the steep rises (equivalence points): one at 20 mL (about pH 3.7 to 5.8) and one at 40 mL (about pH 8 to 11). Two equivalence points, equally spaced: two acidic protons.

CED 8.5.A.5 · Read this in Fix

1. AGraph 1
   A student who thinks a 0.10 M weak acid behaves like a 0.10 M strong acid picks this. This curve starts at pH 1, as 0.10 M HCl would; CH₃COOH is only slightly ionized and starts near pH 3.
2. BGraph 2
   A student who thinks every equivalence point is at pH 7 picks this. At the equivalence point the solution contains CH₃COO⁻, which reacts with water to produce OH⁻, so the pH there is about 8.7.
3. CGraph 3 Correct
   This curve starts near pH 3, rises gradually, has its equivalence point at 25.0 mL (2.5 mmol of CH₃COOH needs 2.5 mmol of NaOH) and has the equivalence point above pH 7, because CH₃COO⁻ ions accept protons from water.
4. DGraph 4
   A student who thinks a weak acid reacts with only part of the added base picks this, with its equivalence point at 12.5 mL. CH₃COOH reacts quantitatively, so 25.0 mL of 0.10 M NaOH is needed.

Working No calculation needed beyond estimates. Initial pH ≈ ½(4.74 + 1.00) = 2.9 (not 1, so not the strong-acid curve). Equivalence at 25.0 mL (2.5 mmol acid, 0.10 M NaOH), not 12.5 mL. At the equivalence point CH₃COO⁻ (0.050 M) makes the solution basic: pH ≈ 8.7, above 7.

CED 8.5.A.1 · Read this in Fix

1. AThe volume of NaOH added at the equivalence point Correct
   At the equivalence point the moles of NaOH added equal the moles of HA originally present, for a weak acid as for a strong one. Moles of HA = [NaOH] × V_eq, and dividing by the sample volume gives the initial concentration.
2. BThe value of K_a for HA at the temperature used
   A student who thinks the amount of base an acid needs depends on its strength picks this. The equivalence relationship uses moles only: moles of HA = [NaOH] × V_eq.
3. CThe pH of the HA solution before NaOH is added
   A student who thinks the concentration of any acid equals 10^−pH picks this. HA is weak and only slightly ionized, so its [H₃O⁺] is much less than its concentration.
4. DThe volume of NaOH added until the pH reaches 7.00
   A student who thinks the equivalence point is always at pH 7 picks this. For a weak acid the equivalence point is above pH 7, so the volume at pH 7 is slightly less than the equivalence volume.

Working No numerical answer. At the equivalence point moles NaOH added = moles HA: [HA]₀ = (0.1000 M × V_eq)/25.00 mL. The sample volume and [NaOH] are given, so the quantity still needed is the volume of NaOH added at the equivalence point. K_a is not needed (the relationship holds for weak acids), the initial pH alone does not give [HA] for a weak acid, and the volume at pH 7.00 is not the equivalence volume for a weak acid.

CED 8.5.A.2 · Read this in Fix

1. A0.250 M
   A student who thinks the equivalence point is where HA and A⁻ are equal reads 12.5 mL, the middle of the flat region, and picks this. That is the half-equivalence point; the equivalence point is at 25.0 mL.
2. B0.200 M
   A student who thinks the analyte and titrant have equal concentrations at the equivalence point picks this. The MOLES are equal: 5.00 × 10⁻³ mol of NaOH reacted with the HA in 10.0 mL.
3. C0.143 M
   A student who divides the moles of HA by the total volume at the equivalence point (35.0 mL) picks this. The initial concentration uses the volume of the original 10.0 mL sample.
4. D0.500 M Correct
   The equivalence point, the middle of the steep rise, is at 25.0 mL. Moles of NaOH = 0.200 M × 0.0250 L = 5.00 × 10⁻³ mol, which equals the moles of HA in the 10.0 mL sample: [HA] = 5.00 × 10⁻³ mol/0.0100 L = 0.500 M.

Working Equivalence point (middle of the steep rise) at 25.0 mL. n(NaOH) = 0.200 M × 0.0250 L = 5.00 × 10⁻³ mol = n(HA). [HA]₀ = 5.00 × 10⁻³ mol/0.0100 L = 0.500 M.

CED 8.5.A.2 · Read this in Fix

1. AIt increases, because the flask now holds a larger volume of solution
   A student who thinks the volume of titrant depends on the volume of analyte solution picks this. The equivalence point depends on moles of HA, which the water does not change.
2. BIt is unchanged, because the moles of HA in the flask are unchanged Correct
   Adding water changes the volume and concentration of the solution in the flask but not the number of moles of HA. The equivalence point is reached when moles of NaOH added equal moles of HA, so the same volume of NaOH is needed.
3. CIt decreases, because the HA in the flask is now more dilute
   A student who thinks diluting an acid reduces the amount of acid picks this. The HA is more dilute, but the moles of HA are the same, so the same moles of NaOH are needed.
4. DIt is unchanged, because adding water does not alter the pH
   A student who thinks dilution leaves the pH of an acid solution unchanged picks this. Adding water does raise the pH of the HA solution; the volume is unchanged because the moles of HA are unchanged.

Working No calculation. Moles HA in the flask are the same in both titrations (water adds none). At the equivalence point moles NaOH = moles HA, so the same volume of the same NaOH solution is needed.

CED 8.5.A.2 · Read this in Fix

1. AIt is higher by about one pH unit
   A student who applies the strong-acid dilution rule, one pH unit per ten-fold dilution, picks this. At the half-equivalence point pH = pK_a, which does not depend on the dilution.
2. BIt is higher by about half a pH unit
   A student who treats the half-equivalence solution as a solution of HA alone picks this, using pH ≈ ½(pK_a − log[HA]). The solution contains equal amounts of HA and A⁻, so pH = pK_a.
3. CIt is unchanged from the first titration Correct
   At the half-equivalence point [HA] = [A⁻], so pH = pK_a. Diluting the acid and the base ten-fold changes both concentrations equally, and pK_a is a constant at a given temperature, so the pH at the half-equivalence point is still 4.70.
4. DIt is lower than in the first titration
   A student who thinks a more dilute weak acid has a larger K_a picks this. The percent ionization of HA alone rises on dilution, but K_a and pK_a stay the same.

Working At the half-equivalence point [HA] = [A⁻], so pH = pK_a + log(1) = pK_a. K_a does not depend on concentration at a fixed temperature, so the pH at the half-equivalence point is 4.70 in both titrations (a full calculation gives 4.70 and 4.71).

CED 8.5.A.3 · Read this in Fix

1. AEqual volumes of NaOH; an equivalence-point pH of 7.00 for both
   A student who thinks every equivalence point is at pH 7 picks this. The volumes are equal, but CH₃COO⁻ present at the CH₃COOH equivalence point makes that solution basic.
2. BMore NaOH for HCl; a higher equivalence-point pH for CH₃COOH
   A student who thinks a stronger acid needs more base picks this. The amount of base needed depends on the moles of acid, which are equal (2.50 mmol each).
3. CMore NaOH for HCl; a lower equivalence-point pH for CH₃COOH
   A student who thinks a weak acid reacts with only part of the base picks this, expecting less NaOH for CH₃COOH and un-neutralized acid at its equivalence point. CH₃COOH reacts completely with OH⁻; its equivalence point is basic.
4. DEqual volumes of NaOH; a higher equivalence-point pH for CH₃COOH Correct
   Each sample contains 2.50 mmol of a monoprotic acid, so each needs 2.50 mmol (25.00 mL) of NaOH. At the equivalence point the HCl titration contains only Na⁺ and Cl⁻ (pH 7.00), while the CH₃COOH titration contains CH₃COO⁻, which accepts protons from water (pH about 8.7).

Working Both samples contain 2.50 mmol of monoprotic acid, so both need 25.00 mL of NaOH. At the HCl equivalence point only Na⁺ and Cl⁻ remain: pH 7.00. At the CH₃COOH equivalence point CH₃COO⁻ (0.0500 M) reacts with water: pH ≈ 8.72. So: equal volumes; higher pH for CH₃COOH.

CED 8.5.A.2 · Read this in Fix

1. AHY is the stronger acid, because its pH at half-equivalence is higher
   A student who thinks a higher pK_a means a stronger acid picks this. pK_a = −log K_a, so HY's higher pK_a (5.20) means a smaller K_a and a weaker acid.
2. BHX is the stronger acid, because its pH at half-equivalence is lower Correct
   The pH at the half-equivalence point equals pK_a: 3.80 for HX and 5.20 for HY. A lower pK_a means a larger K_a, so HX is the stronger acid.
3. CHY is the stronger acid, because its pH at equivalence is higher
   A student who thinks a stronger acid gives a more basic equivalence point picks this. The higher equivalence-point pH of HY shows that Y⁻ is the stronger base, which means HY is the weaker acid.
4. DThey are equally strong, because they need the same volume of NaOH
   A student who thinks acid strength decides how much base is needed picks this. Equal volumes show only that the samples contained equal moles of acid; strength is shown by pK_a.

Working pK_a = pH at half-equivalence: HX 3.80, HY 5.20. Lower pK_a means larger K_a: K_a(HX) = 1.6 × 10⁻⁴ > K_a(HY) = 6.3 × 10⁻⁶. HX is the stronger acid. (Equal volumes reflect equal moles; the higher equivalence pH of HY reflects its stronger conjugate base.)

CED 8.5.A.3 · Read this in Fix

1. ACH₃COOH(aq) + H₂O(l) ⇌ CH₃COO⁻(aq) + H₃O⁺(aq)
   A student who thinks a weak acid reacts with only part of the base, leaving CH₃COOH at the equivalence point, picks this. CH₃COOH reacts completely with OH⁻, so almost none remains.
2. BNa⁺(aq) + 2 H₂O(l) ⇌ NaOH(aq) + H₃O⁺(aq)
   A student who thinks Na⁺ ions react with water picks this. Na⁺ is a spectator ion and does not react with water to any significant extent.
3. CH₂O(l) + H₂O(l) ⇌ H₃O⁺(aq) + OH⁻(aq)
   A student who thinks the equivalence point is neutral, so that only water's autoionization matters, picks this. CH₃COO⁻ produces far more OH⁻ than water's autoionization does.
4. DCH₃COO⁻(aq) + H₂O(l) ⇌ CH₃COOH(aq) + OH⁻(aq) Correct
   At the equivalence point the CH₃COOH has been converted to CH₃COO⁻. This conjugate base accepts protons from water, producing OH⁻, so this equilibrium sets the (basic) pH.

Working No calculation. At the equivalence point all CH₃COOH has been converted to CH₃COO⁻; Na⁺ is a spectator. The major species that reacts with water is CH₃COO⁻: CH₃COO⁻ + H₂O ⇌ CH₃COOH + OH⁻, making the solution basic.

CED 8.5.A.4 · Read this in Fix

1. A6.5 Correct
   The equivalence points are at 20 mL and 40 mL. The second proton is removed between them, so its half-equivalence point is at 30 mL, where [HA⁻] = [A²⁻]; the curve shows pH 6.5 there, so pK_a for the second proton is 6.5.
2. B3.0
   A student who thinks a polyprotic acid has a single pK_a, read at the first half-equivalence point (10 mL), picks this. That value is pK_a for the first proton; the second has its own pK_a.
3. C4.8
   A student who takes half of the total volume to the last equivalence point (20 mL) as the second half-equivalence point picks this. 20 mL is the first equivalence point; the second proton's half-equivalence point is at 30 mL.
4. D9.5
   A student who thinks pK_a is the pH at an equivalence point reads the pH at 40 mL and picks this. pK_a is read halfway to that equivalence point, at 30 mL.

Working First equivalence point at 20.0 mL, second at 40.0 mL. The second proton is removed between them; halfway, at 30.0 mL, [HA⁻] = [A²⁻] and pH = pK_a2 = 6.5 (read from the curve).

CED 8.5.A.5 · Read this in Fix

1. AH₂A
   A student who thinks a weak acid reacts with only part of the added base picks this. H₂A reacts quantitatively with OH⁻, so by the first equivalence point almost all of it has become HA⁻.
2. BA²⁻
   A student who thinks a diprotic acid loses both protons at once picks this. The protons are removed one at a time; A²⁻ becomes the major species only at the second equivalence point.
3. COH⁻
   A student who thinks OH⁻ is a major species in any basic solution picks this. At pH 8.5, [OH⁻] = 10^−5.5 ≈ 3 × 10⁻⁶ M, far less than the concentration of HA⁻.
4. DHA⁻ Correct
   At the first equivalence point the OH⁻ added has removed one proton from every H₂A molecule, so HA⁻ is the major species. Only small amounts of H₂A and A²⁻ form from HA⁻, and at pH 8.5 [OH⁻] is about 3 × 10⁻⁶ M.

Working No calculation. Protons are removed one at a time: by the first equivalence point the moles of OH⁻ added equal the moles of H₂A, converting H₂A to HA⁻. HA⁻ is the major species; H₂A and A²⁻ are minor, and [OH⁻] at pH 8.5 is only about 3 × 10⁻⁶ M.

CED 8.5.A.5 · Read this in Fix

1. A1.0 × 10⁻⁸
   A student who thinks the pH at the half-equivalence point equals pK_b of the base picks this, taking pK_b = 8.0. That pH is pK_a of the conjugate acid HB⁺; pK_b = 14.0 − 8.0 = 6.0.
2. B1.0 × 10⁻⁶ Correct
   The equivalence point is at 25.0 mL, so the half-equivalence point is at 12.5 mL, where pH = 8.0. There [B] = [HB⁺], so pK_a of HB⁺ = 8.0, pK_b = 14.0 − 8.0 = 6.0, and K_b = 1.0 × 10⁻⁶.
3. C2.2 × 10⁻⁵
   A student who reads the pH at the equivalence point (about 4.65) and takes it as pK_b picks this. The constant is read at the half-equivalence point, 12.5 mL.
4. D3.2 × 10⁻⁴
   A student who takes pK_b to be the pOH of the base before any acid is added (14.0 − 10.5 = 3.5) picks this. That pOH also depends on the concentration of B; the half-equivalence point gives the constant.

Working Equivalence point at 25.0 mL; half-equivalence at 12.5 mL, pH 8.0. There [B] = [HB⁺], so pH = pK_a(HB⁺) = 8.0. pK_b = 14.0 − 8.0 = 6.0; K_b = 1.0 × 10⁻⁶.

CED 8.5.A.3 · Read this in Fix

1. A122 g/mol
   A student who thinks the amounts of HA and A⁻ are equal at the equivalence point concludes that the 2.50 × 10⁻³ mol of NaOH has reacted with only half of the acid, takes the sample as 5.00 × 10⁻³ mol, and gets 0.610 g/5.00 × 10⁻³ mol = 122 g/mol. At the equivalence point all of the HA has reacted, so the sample contained 2.50 × 10⁻³ mol.
2. B203 g/mol
   A student who thinks the analyte solution has the same concentration as the titrant at the equivalence point takes the 30.0 mL of acid solution as 0.100 M, 3.00 × 10⁻³ mol, and gets 0.610 g/3.00 × 10⁻³ mol = 203 g/mol. It is the moles of titrant and analyte that are equal, and the moles of NaOH are 0.100 M × 0.02500 L = 2.50 × 10⁻³ mol.
3. C244 g/mol Correct
   The NaOH delivered is 26.20 − 1.20 = 25.00 mL, which contains 0.100 M × 0.02500 L = 2.50 × 10⁻³ mol. At the equivalence point this equals the moles of HA in the sample, so the molar mass is 0.610 g/2.50 × 10⁻³ mol = 244 g/mol.
4. D447 g/mol
   A student who takes the concentration of the acid solution to be its moles divided by the total volume in the flask at the equivalence point calculates 2.50 × 10⁻³ mol/0.0550 L = 0.0455 M, multiplies by the 30.0 mL of acid solution to get 1.36 × 10⁻³ mol, and finds 0.610 g/1.36 × 10⁻³ mol = 447 g/mol. The moles of HA in the sample equal the moles of NaOH added, 2.50 × 10⁻³ mol, with no volume correction.

Working Volume of NaOH added at the equivalence point = 26.20 mL − 1.20 mL = 25.00 mL = 0.02500 L. Moles of NaOH = 0.100 M × 0.02500 L = 2.50 × 10⁻³ mol. At the equivalence point moles of titrant = moles of monoprotic acid originally present, so moles of HA = 2.50 × 10⁻³ mol. Molar mass = 0.610 g/2.50 × 10⁻³ mol = 244 g/mol. The volume of water used to dissolve the acid is not needed.

CED 8.5.A.2 · Read this in Fix