8.4 Acid-Base Reactions and Buffers

1. ADiagram 1
   A student who thinks the salt formed in a neutralization exists as NaCl units picks this. Na⁺ and Cl⁻ are spectator ions and stay as separate hydrated ions; the correct diagram shows 3 separate Cl⁻ ions and 2 separate Na⁺ ions.
2. BDiagram 2
   A student who thinks the ions of two solutions simply spread through the mixture without reacting picks this. H₃O⁺ and OH⁻ react to form water, so the 2 OH⁻ ions are used up along with 2 of the 3 H₃O⁺ ions.
3. CDiagram 3 Correct
   H₃O⁺ and OH⁻ react in a 1:1 ratio: the 2 OH⁻ ions react with 2 of the 3 H₃O⁺ ions to form water, leaving 1 H₃O⁺ ion. Na⁺ and Cl⁻ do not react and remain as separate ions: 3 Cl⁻ and 2 Na⁺.
4. DDiagram 4
   A student who thinks mixing an acid with a base always gives a neutral solution picks this. There were 3 H₃O⁺ ions and only 2 OH⁻ ions, so one H₃O⁺ ion is left over and the mixture is acidic.

Working No calculation. Before mixing: 3 H₃O⁺ and 3 Cl⁻; 2 Na⁺ and 2 OH⁻. H₃O⁺ + OH⁻ → 2 H₂O uses 2 H₃O⁺ and both OH⁻, leaving 1 H₃O⁺; Na⁺ and Cl⁻ are spectators and stay as separate ions. After: 1 H₃O⁺, 3 Cl⁻, 2 Na⁺.

CED 8.4.A.1 · Read this in Fix

1. A6.0 × 10⁻² M
   A student who thinks a weak acid reacts with only a small fraction of a strong base picks this, keeping nearly all 6.0 mmol of OH⁻. HA reacts quantitatively with OH⁻, so 4.0 mmol of OH⁻ are used up and only 2.0 mmol remain.
2. B2.0 × 10⁻² M Correct
   The graph gives 4.0 mmol of HA and 6.0 mmol of OH⁻. HA + OH⁻ → A⁻ + H₂O goes to completion, leaving 2.0 mmol of OH⁻ in 100.0 mL: [OH⁻] = 2.0 × 10⁻² M. The OH⁻ produced by A⁻ reacting with water is negligible beside this excess.
3. C3.3 × 10⁻² M
   A student who divides the excess OH⁻ by the volume of the NaOH solution alone picks this: 2.0 mmol/60.0 mL. The excess OH⁻ is spread through the whole 100.0 mL of the mixture.
4. D6.3 × 10⁻⁶ M
   A student who always takes the pH-determining reaction to be A⁻ + H₂O ⇌ HA + OH⁻ picks this, using [A⁻] = 0.040 M and K_b = 1.0 × 10⁻⁹. Here the strong base is in excess, so the 2.0 mmol of excess OH⁻ dominates.

Working From the graph: n(HA) = 4.0 mmol, n(OH⁻) = 6.0 mmol. HA + OH⁻ → A⁻ + H₂O is quantitative: 4.0 mmol HA use 4.0 mmol OH⁻, leaving 2.0 mmol OH⁻ (and 4.0 mmol A⁻). Total volume 40.0 + 60.0 = 100.0 mL. [OH⁻] = 2.0 mmol/100.0 mL = 2.0 × 10⁻² M. (OH⁻ from A⁻ + H₂O is about 6 × 10⁻⁶ M, negligible.)

CED 8.4.A.2 · Read this in Fix

1. ANH₃(aq) + HCl(aq) → NH₄Cl(aq)
   A student who thinks HCl reacts as molecules to form the salt picks this. HCl is completely ionized in water, and NH₄⁺ and Cl⁻ remain as separate ions, so neither HCl nor NH₄Cl appears as a particle in the solution.
2. BOH⁻(aq) + H⁺(aq) → H₂O(l)
   A student who thinks every neutralization is H⁺ + OH⁻ → H₂O picks this. NH₃ is a weak base that is mostly un-ionized, so the species that accepts the proton is the NH₃ molecule, not OH⁻.
3. CNH₃(aq) + H⁺(aq) → NH₄⁺(aq) Correct
   In the mixture, HCl is completely ionized and NH₃ is present mostly as molecules. The reaction is the transfer of a proton from H⁺ (H₃O⁺) to NH₃, forming NH₄⁺; Cl⁻ is a spectator ion and is left out of the net ionic equation.
4. DCl⁻(aq) + NH₄⁺(aq) → NH₄Cl(aq)
   A student who thinks the salt forms as NH₄Cl units in solution picks this. NH₄⁺ and Cl⁻ stay as separate hydrated ions; the reaction that occurs is the proton transfer to NH₃.

Working No calculation. HCl is a strong acid: HCl(aq) is H⁺ (H₃O⁺) and Cl⁻ ions. NH₃ is a weak base, present mostly as NH₃ molecules. The proton transfers from H⁺ to NH₃: NH₃(aq) + H⁺(aq) → NH₄⁺(aq); Cl⁻ is a spectator.

CED 8.4.A.3 · Read this in Fix

1. ADiagram 1
   A student who thinks reactants and products are present in equal amounts at equilibrium picks this. Two of each species gives K = (2 × 2)/(2 × 2) = 1, not 9.0.
2. BDiagram 2
   A student who thinks every acid-base reaction goes to completion picks this. A weak acid and a weak base react to an equilibrium state; with no HA or B left, the mixture could not have K = 9.0.
3. CDiagram 3 Correct
   Each diagram began with four HA and four B. In this diagram 1 HA, 1 B, 3 A⁻ and 3 HB⁺ are present, so K = (3 × 3)/(1 × 1) = 9.0, the given value. The mixture contains both reactants and products, as expected for an equilibrium of a weak acid with a weak base.
4. DDiagram 4
   A student who thinks a weak acid and a weak base hardly react with each other picks this. With K = 9.0, products are favored; a mixture with no A⁻ or HB⁺ is not at equilibrium.

Working Each diagram has four acid units and four base units in total. With equal volumes for all species, K = (n_A⁻ × n_HB⁺)/(n_HA × n_B). 1 HA, 1 B, 3 A⁻, 3 HB⁺: K = (3 × 3)/(1 × 1) = 9.0 ✓. 2, 2, 2, 2: K = 1. 0 HA, 0 B: K undefined (complete reaction). 4 HA, 4 B, no products: K = 0.

CED 8.4.A.4 · Read this in Fix

1. A7.00
   A student who thinks mixing an acid with a base always gives a neutral solution picks this. The HCl is in excess (3.0 mmol of H₃O⁺ against 2.0 mmol of OH⁻), so 1.0 mmol of H₃O⁺ remains and the solution is acidic.
2. B1.48
   A student who divides the excess moles by the volume of the HCl solution alone picks this: 1.0 mmol/30.0 mL = 0.033 M. After mixing, the excess H₃O⁺ is spread through 50.0 mL, so [H₃O⁺] = 0.020 M.
3. C5.80
   A student who averages the pH values of the two solutions, weighted by volume, picks this: (30.0 × 1.00 + 20.0 × 13.00)/50.0 = 5.80. pH values do not average; H₃O⁺ and OH⁻ react, and the 1.0 mmol of H₃O⁺ left over gives pH 1.70.
4. D1.70 Correct
   3.0 mmol of H₃O⁺ react with 2.0 mmol of OH⁻, leaving 1.0 mmol of H₃O⁺ in the total volume of 50.0 mL: [H₃O⁺] = 1.0 mmol/50.0 mL = 0.020 M, so pH = −log(0.020) = 1.70.

Working n(H₃O⁺) = 0.0300 L × 0.10 M = 3.0 × 10⁻³ mol; n(OH⁻) = 0.0200 L × 0.10 M = 2.0 × 10⁻³ mol. H₃O⁺ + OH⁻ → 2 H₂O leaves 1.0 × 10⁻³ mol H₃O⁺ in 0.0500 L: [H₃O⁺] = 0.020 M. pH = −log(0.020) = 1.70.

CED 8.4.A.1 · Read this in Fix

1. AUse pK_a of HA and the ratio of the moles of A⁻ to the moles of HA Correct
   HA (4.0 mmol) is in excess over OH⁻ (2.0 mmol); after reaction the mixture contains 2.0 mmol HA and 2.0 mmol A⁻, the buffer case of the EK. Its pH comes from pK_a and the ratio [A⁻]/[HA], which equals the ratio of moles because both are in the same volume.
2. BUse K_b of A⁻ and the concentration of A⁻ in the mixture
   A student who thinks the A⁻ + H₂O equilibrium always sets the pH after a weak acid reacts with a strong base picks this. That method applies only to an equimolar mixture; here 2.0 mmol of HA remain alongside the A⁻.
3. CUse the moles of OH⁻ added and the total volume of the mixture
   A student who thinks a weak acid reacts with only a small fraction of added OH⁻ picks this, expecting the OH⁻ to remain. HA reacts quantitatively, and the 2.0 mmol of OH⁻ are all used up, so no excess OH⁻ remains.
4. DUse K_a of HA and the concentration of HA before it was mixed
   A student who applies the weak-acid-alone method to any solution containing a weak acid picks this. The NaOH has converted half of the HA to A⁻, so the solution is no longer a solution of HA alone.

Working No numerical answer needed. n(HA) = 4.0 mmol, n(OH⁻) = 2.0 mmol; after HA + OH⁻ → A⁻ + H₂O: 2.0 mmol HA and 2.0 mmol A⁻, no excess OH⁻. The weak acid was in excess, so the mixture is a buffer and pH = pK_a + log([A⁻]/[HA]), where the ratio of concentrations equals the ratio of moles (here 1, so pH = pK_a).

CED 8.4.A.2 · Read this in Fix

1. AMix 100.0 mL of the HA solution with 100.0 mL of the NaOH solution
   A student who thinks mixing equal amounts of weak acid and strong base makes a buffer picks this. The 10 mmol of each react completely, leaving A⁻ with almost no HA, so the pH is set by A⁻ reacting with water.
2. BMix 20.0 mL of the HA solution with 60.0 mL of the NaOH solution
   A student who thinks any mixture of a weak acid and a strong base is a buffer picks this. Here OH⁻ (6.0 mmol) is in excess over HA (2.0 mmol): all the HA is used up and the excess OH⁻ sets the pH.
3. CMix 30.0 mL of the HA solution with 30.0 mL of distilled water
   A student who thinks a solution of a weak acid alone is a buffer picks this. Diluting HA leaves a solution of mostly HA with very little A⁻; mixing with a smaller amount of strong base is what gives both HA and A⁻.
4. DMix 100.0 mL of the HA solution with 40.0 mL of the NaOH solution Correct
   10 mmol of HA react with 4.0 mmol of OH⁻, leaving 6.0 mmol of HA and forming 4.0 mmol of A⁻. The weak acid is in excess, which is the condition the EK gives for a buffer solution to form.

Working No calculation needed beyond amounts. (A buffer forms when the weak acid is in excess.) 100.0 mL HA + 40.0 mL NaOH: 10 mmol HA, 4.0 mmol OH⁻ → 6.0 mmol HA + 4.0 mmol A⁻: buffer. 100.0 + 100.0 mL: 10 mmol each → only A⁻: not a buffer. 20.0 mL HA + 60.0 mL NaOH: 2.0 mmol HA, 6.0 mmol OH⁻ → excess OH⁻: not a buffer. HA + water: only HA (very little A⁻): not a buffer.

CED 8.4.A.2 · Read this in Fix

1. A9.20 Correct
   Equal amounts (10 mmol) react, giving 10 mmol of A⁻ in 100.0 mL, so [A⁻] = 0.10 M. K_b = K_w/K_a = 2.5 × 10⁻⁹, so [OH⁻] = √(2.5 × 10⁻⁹ × 0.10) = 1.6 × 10⁻⁵ M, pOH = 4.80 and pH = 9.20.
2. B3.05
   A student who finds the pH of any solution containing a weak acid from K_a and the initial acid concentration picks this: √(4.0 × 10⁻⁶ × 0.20) = 8.9 × 10⁻⁴ M. All the HA has reacted with the NaOH, so the solution contains A⁻, not HA.
3. C7.00
   A student who thinks equal amounts of an acid and a base always give a neutral solution picks this. After reaction the solution contains A⁻, a weak base, which reacts with water to produce OH⁻, so the pH is above 7.
4. D9.35
   A student who uses [A⁻] = 0.20 M, the concentration of the original solutions, picks this: [OH⁻] = √(2.5 × 10⁻⁹ × 0.20) = 2.2 × 10⁻⁵ M. Mixing doubles the volume, so [A⁻] = 0.10 M and [OH⁻] = 1.6 × 10⁻⁵ M.

Working n(HA) = n(OH⁻) = 10 mmol: equimolar. After HA + OH⁻ → A⁻ + H₂O: 10 mmol A⁻ in 100.0 mL, [A⁻] = 0.10 M. K_b = K_w/K_a = 1.0 × 10⁻¹⁴/4.0 × 10⁻⁶ = 2.5 × 10⁻⁹. A⁻ + H₂O ⇌ HA + OH⁻: [OH⁻] ≈ √(2.5 × 10⁻⁹ × 0.10) = 1.6 × 10⁻⁵ M; pOH = 4.80; pH = 14.00 − 4.80 = 9.20.

CED 8.4.A.2 · Read this in Fix

1. AIt is consistent, since NH₃ is a weak base and reacts with only part of the H₃O⁺
   A student who thinks a weak base reacts with only a small fraction of a strong acid picks this. NH₃ ionizes only slightly in water, but it reacts quantitatively with H₃O⁺, so very little NH₃ and H₃O⁺ remain.
2. BIt is not consistent, since NH₃ and H₃O⁺ react until almost none of either remains Correct
   The student's diagram shows two NH₃ molecules and two H₃O⁺ ions left over. NH₃ reacts quantitatively with H₃O⁺, and equal amounts were mixed, so after mixing the solute particles should be almost entirely NH₄⁺ and Cl⁻ in equal numbers.
3. CIt is consistent, since at equilibrium reactants and products are present in equal amounts
   A student who thinks equilibrium means equal amounts of reactants and products picks this. The reaction of NH₃ with H₃O⁺ goes essentially to completion, so the products NH₄⁺ and Cl⁻ greatly outnumber any remaining NH₃ and H₃O⁺.
4. DIt is not consistent, since NH₄⁺ and Cl⁻ should be drawn joined as NH₄Cl units
   A student who thinks the salt exists as NH₄Cl units in solution picks this. NH₄⁺ and Cl⁻ are separate ions in water; the error in the diagram is the unreacted NH₃ and H₃O⁺.

Working No calculation. Equal moles of NH₃ and H₃O⁺. NH₃ + H₃O⁺ → NH₄⁺ + H₂O is quantitative, so after mixing almost all NH₃ and H₃O⁺ are used up: the particles should be NH₄⁺ and Cl⁻ in equal numbers (4 each). The small amount of NH₃ and H₃O⁺ re-formed by NH₄⁺ + H₂O ⇌ NH₃ + H₃O⁺ is far too small to show as two of each.

CED 8.4.A.3 · Read this in Fix

1. A1.26
   A student who thinks a weak base reacts with only a small fraction of a strong acid picks this, keeping nearly all 2.5 mmol of H₃O⁺. NH₃ reacts quantitatively, leaving only 0.50 mmol of H₃O⁺.
2. B1.70
   A student who divides the excess H₃O⁺ by the volume of the HCl solution alone picks this: 0.50 mmol/25.0 mL. The excess is spread through the total volume of 45.0 mL.
3. C5.30
   A student who thinks the NH₄⁺ + H₂O equilibrium always sets the pH after a weak base reacts with a strong acid picks this, using [NH₄⁺] = 0.044 M and K_a = 5.6 × 10⁻¹⁰. The strong acid is in excess, so the excess H₃O⁺ sets the pH.
4. D1.95 Correct
   2.0 mmol of NH₃ react with 2.0 mmol of the 2.5 mmol of H₃O⁺, leaving 0.50 mmol of H₃O⁺ in 45.0 mL: [H₃O⁺] = 0.011 M, pH = 1.95. The H₃O⁺ produced by NH₄⁺ reacting with water is negligible beside this excess of strong acid.

Working n(NH₃) = 2.0 mmol; n(H₃O⁺) = 2.5 mmol. NH₃ + H₃O⁺ → NH₄⁺ + H₂O leaves 0.50 mmol H₃O⁺ (and 2.0 mmol NH₄⁺) in 45.0 mL. [H₃O⁺] = 0.50/45.0 = 0.011 M; pH = 1.95. (H₃O⁺ from NH₄⁺, K_a = 5.6 × 10⁻¹⁰, is about 5 × 10⁻⁶ M: negligible.)

CED 8.4.A.3 · Read this in Fix

1. AThe HY mixture, because Y⁻ is a stronger base than X⁻ Correct
   Each mixture is equimolar, so the pH is set by the conjugate base reacting with water. K_b = K_w/K_a: 1.0 × 10⁻⁶ for Y⁻ and 1.0 × 10⁻¹⁰ for X⁻. Y⁻ produces more OH⁻, so the HY mixture has the higher pH (about 10.4 against 8.4).
2. BThe HX mixture, because X⁻ is the stronger of the two bases
   A student who thinks a stronger acid has a stronger conjugate base picks this. HX is the stronger acid, so its conjugate base X⁻ is the weaker base: K_b(X⁻) = 1.0 × 10⁻¹⁰, smaller than K_b(Y⁻) = 1.0 × 10⁻⁶.
3. CNeither, because the acid and base in each are equal in amount
   A student who thinks equal amounts of an acid and a base always give a neutral solution picks this. Each mixture contains a conjugate base that reacts with water, so both are basic, and to different extents.
4. DThe HY mixture, because HY leaves most of the added OH⁻ unreacted
   A student who thinks a weak acid reacts with only a small fraction of an added strong base, and the weaker acid HY with the least, picks this. Both acids react essentially completely with OH⁻ (more than 99% of it in each mixture); the HY mixture has the higher pH because Y⁻, the stronger conjugate base, reacts with water to a greater extent.

Working Both are equimolar (2.5 mmol each), giving 0.050 M X⁻ or Y⁻. K_b(X⁻) = 1.0 × 10⁻¹⁴/1.0 × 10⁻⁴ = 1.0 × 10⁻¹⁰; K_b(Y⁻) = 1.0 × 10⁻⁶. [OH⁻] ≈ √(K_b × 0.050): X⁻ 2.2 × 10⁻⁶ M (pH 8.35); Y⁻ 2.2 × 10⁻⁴ M (pH 10.35). The HY mixture has the higher pH because Y⁻ is the stronger base.

CED 8.4.A.2 · Read this in Fix

1. AReactants are favored, because NH₄⁺ is a stronger acid than CH₃COOH
   A student who thinks the conjugate acid of a weak base is a strong acid picks this. NH₄⁺ has K_a = K_w/K_b = 5.6 × 10⁻¹⁰, so it is a much weaker acid than CH₃COOH.
2. BProducts are favored, because CH₃COOH is a stronger acid than NH₄⁺ Correct
   K_a of NH₄⁺ = K_w/K_b = 5.6 × 10⁻¹⁰, far smaller than K_a of CH₃COOH, 1.8 × 10⁻⁵. The proton moves from the stronger acid to form the weaker acid, so products are favored: K = (1.8 × 10⁻⁵)(1.8 × 10⁻⁵)/(1.0 × 10⁻¹⁴) = 3.2 × 10⁴.
3. CReactants are favored, because a weak acid and a weak base barely react
   A student who thinks two weak species hardly react picks this. The extent of reaction depends on K = K_a × K_b/K_w, here 3.2 × 10⁴, so products are strongly favored.
4. DNeither is favored, because the K_a of CH₃COOH equals the K_b of NH₃
   A student who compares K_a of the acid directly with K_b of the base picks this. K = K_a × K_b/K_w = 3.2 × 10⁴, not 1; the comparison that matters is between CH₃COOH and NH₄⁺ as acids.

Working K_a(NH₄⁺) = K_w/K_b(NH₃) = 1.0 × 10⁻¹⁴/1.8 × 10⁻⁵ = 5.6 × 10⁻¹⁰. CH₃COOH (K_a = 1.8 × 10⁻⁵) is a much stronger acid than NH₄⁺, so the proton transfer to NH₃ is favored. K = K_a × K_b/K_w = (1.8 × 10⁻⁵)²/1.0 × 10⁻¹⁴ = 3.2 × 10⁴ > 1: products favored.

CED 8.4.A.4 · Read this in Fix

1. A0.080 M
   A student who thinks a weak base reacts with only a small fraction of an added strong acid treats nearly all 4.0 × 10⁻³ mol of NH₃ as still present in the 50.0 mL and gets 0.080 M. The reaction of NH₃ with H₃O⁺ goes essentially to completion, so 1.0 × 10⁻³ mol of NH₃ is used up and 0.060 M remains.
2. B0.075 M
   A student who divides the moles of excess base by the volume of the solution it came from calculates 3.0 × 10⁻³ mol/0.0400 L = 0.075 M. After mixing, the NH₃ is spread through the total volume, 50.0 mL, which gives 0.060 M.
3. C0.060 M Correct
   The 1.0 × 10⁻³ mol of H₃O⁺ reacts essentially completely with NH₃, leaving 4.0 × 10⁻³ − 1.0 × 10⁻³ = 3.0 × 10⁻³ mol of NH₃ in a total volume of 50.0 mL: 3.0 × 10⁻³ mol/0.0500 L = 0.060 M.
4. D0.040 M
   A student who thinks reactants and products are present in equal amounts at equilibrium splits the 4.0 × 10⁻³ mol into 2.0 × 10⁻³ mol of NH₃ and 2.0 × 10⁻³ mol of NH₄⁺ and gets 2.0 × 10⁻³ mol/0.0500 L = 0.040 M. The amounts are set by the stoichiometry: only 1.0 × 10⁻³ mol of H₃O⁺ was added, so 3.0 × 10⁻³ mol of NH₃ remains.

Working Moles of NH₃ = 0.0400 L × 0.10 M = 4.0 × 10⁻³ mol; moles of H₃O⁺ = 0.0100 L × 0.10 M = 1.0 × 10⁻³ mol. NH₃(aq) + H₃O⁺(aq) → NH₄⁺(aq) + H₂O(l) goes essentially to completion, and H₃O⁺ is limiting, so 4.0 × 10⁻³ − 1.0 × 10⁻³ = 3.0 × 10⁻³ mol of NH₃ remains (with 1.0 × 10⁻³ mol of NH₄⁺). Total volume = 40.0 mL + 10.0 mL = 50.0 mL = 0.0500 L. [NH₃] = 3.0 × 10⁻³ mol/0.0500 L = 0.060 M. (The further ionization of the remaining NH₃ changes this by less than 0.1%.)

CED 8.4.A.3 · Read this in Fix

1. AThe volume of the NaOH solution before mixing
   A student who thinks the excess base has the concentration given by its moles divided by the volume of the solution it came from asks for the NaOH volume. The excess OH⁻ is spread through the whole mixture, so the total volume is needed.
2. BThe total volume of the mixture after mixing Correct
   With the strong base in excess, the pH is found from the moles of excess OH⁻ (moles of NaOH minus moles of HA) and the total volume of the solution, which together give [OH⁻].
3. CThe K_b value for A⁻, the conjugate base of HA
   A student who thinks the pH after a weak acid reacts with a strong base is always set by the reaction of A⁻ with water asks for K_b. That equilibrium sets the pH only when the amounts are equimolar; here the excess OH⁻ from NaOH is far greater than the OH⁻ that A⁻ produces.
4. DThe K_a value for HA, the acid that was mixed in
   A student who thinks the pH of any solution made from a weak acid is found from K_a and the acid's initial concentration asks for K_a. Essentially all the HA has been converted to A⁻, and the pH is set by the OH⁻ left over.

Working HA(aq) + OH⁻(aq) → A⁻(aq) + H₂O(l) goes essentially to completion, so with NaOH in excess, moles of OH⁻ left = moles of NaOH − moles of HA. [OH⁻] = moles of excess OH⁻ ÷ total volume of the mixture; then pOH = −log[OH⁻] and pH = 14.00 − pOH at 25°C. The only quantity still missing is the total volume. The OH⁻ produced by A⁻ reacting with water is negligible beside the excess OH⁻, so no K_a or K_b is needed.

CED 8.4.A.2 · Read this in Fix