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AP Chemistry – Units 7–9 Practice Set

Equilibrium, acids and bases, applications of thermodynamics.

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12 questions  ·  51 marks  ·  answers included  ·  AP Chemistry (College Board CED, Units 7–9)
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This worksheet covers:

  • Unit 7 (Equilibrium)
  • Unit 8 (Acids and Bases)
  • Unit 9 (Applications of Thermodynamics)

A calculator is permitted for all questions, in line with the AP Chemistry exam. Use $K_w = 1.0\times10^{-14}$ and $F = 96{,}485\,\text{C mol}^{-1}$ where needed.

Questions

Section A – Equilibrium (Unit 7)

1

At a certain temperature, $0.050\,\text{mol}$ of $\text{N}_2\text{O}_4$ is placed in a $1.00\,\text{L}$ flask and allowed to reach equilibrium: $\text{N}_2\text{O}_4\text{(g)} \rightleftharpoons 2\text{NO}_2\text{(g)}$. At equilibrium, $[\text{NO}_2] = 0.030\,\text{M}$.(a) Determine $[\text{N}_2\text{O}_4]$ at equilibrium.(b) Calculate the value of $K_c$ at this temperature.

Show answer
(a) By stoichiometry, $[\text{N}_2\text{O}_4]_{eq} = 0.050 – \dfrac{0.030}{2} = 0.035\,\text{M}$(b) $K_c = \dfrac{[\text{NO}_2]^2}{[\text{N}_2\text{O}_4]} = \dfrac{(0.030)^2}{0.035} = 2.6\times10^{-2}$ (2 s.f.)
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2

For the reaction $\text{H}_2\text{(g)} + \text{I}_2\text{(g)} \rightleftharpoons 2\text{HI(g)}$, $K_c = 54.3$ at a given temperature. A flask is prepared with $[\text{H}_2]_0 = [\text{I}_2]_0 = 0.200\,\text{M}$ and no HI.(a) Set up an ICE table for this system and, using the fact that the initial concentrations of $\text{H}_2$ and $\text{I}_2$ are equal, show that $\dfrac{x}{0.200-x} = \sqrt{K_c}$.(b) Calculate the equilibrium concentration of HI.

Show answer
(a) ICE: $[\text{H}_2] = [\text{I}_2] = 0.200-x$, $[\text{HI}] = 2x$. Then $K_c = \dfrac{(2x)^2}{(0.200-x)^2} = 54.3$; taking the square root of both sides (valid since all terms are positive) gives $\dfrac{2x}{0.200-x} = \sqrt{54.3}$, i.e. $\dfrac{x}{0.200-x} = \dfrac{\sqrt{54.3}}{2}$ (equivalent rearrangement).(b) Solving, $x = 0.176\,\text{M}$, so $[\text{HI}]_{eq} = 2x = 0.352\,\text{M}$.
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3

The solubility product of lead(II) iodide is $K_{sp} = 7.1\times10^{-9}$ for $\text{PbI}_2\text{(s)} \rightleftharpoons \text{Pb}^{2+}\text{(aq)} + 2\text{I}^{-}\text{(aq)}$.(a) Write the $K_{sp}$ expression for this equilibrium.(b) Calculate the molar solubility of $\text{PbI}_2$ in pure water.

Show answer
(a) $K_{sp} = [\text{Pb}^{2+}][\text{I}^{-}]^2$(b) Let solubility $= s$; then $[\text{Pb}^{2+}] = s$ and $[\text{I}^{-}] = 2s$, so $K_{sp} = s(2s)^2 = 4s^3$. $s = \left(\dfrac{7.1\times10^{-9}}{4}\right)^{1/3} = 1.2\times10^{-3}\,\text{M}$.
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4

For the exothermic reaction $2\text{SO}_2\text{(g)} + \text{O}_2\text{(g)} \rightleftharpoons 2\text{SO}_3\text{(g)}$, predict and explain the effect of each change on the equilibrium position and on the value of $K_c$: (a) decreasing the volume of the container at constant temperature, (b) increasing the temperature.

Show answer
(a) Decreasing volume increases all concentrations/partial pressures; the system shifts toward the side with fewer moles of gas (the product side, 2 mol vs 3 mol), so the equilibrium shifts right, increasing the yield of $\text{SO}_3$. $K_c$ is unchanged, since it depends only on temperature.(b) Increasing temperature shifts the equilibrium left (toward reactants), since the reaction is exothermic and the system favors the endothermic reverse direction to partially absorb the added energy. Because the equilibrium position shifts toward reactants, $K_c$ decreases.
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Section B – Acids and bases (Unit 8)

5

Acetic acid, $\text{CH}_3\text{COOH}$, has $K_a = 1.8\times10^{-5}$. Calculate the pH of a $0.10\,\text{M}$ solution of acetic acid.

Show answer
ICE: $K_a = \dfrac{x^2}{0.10-x} = 1.8\times10^{-5}$. Solving the quadratic (or using the approximation and checking it is valid) gives $x = [\text{H}^+] = 1.33\times10^{-3}\,\text{M}$, so $\text{pH} = -\log(1.33\times10^{-3}) = 2.88$ (3 s.f.).
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6

A buffer is prepared by mixing $0.15\,\text{mol}$ of $\text{CH}_3\text{COOH}$ and $0.25\,\text{mol}$ of $\text{CH}_3\text{COONa}$ in enough water to make $1.00\,\text{L}$ of solution ($K_a = 1.8\times10^{-5}$).(a) Calculate the pH of this buffer using the Henderson–Hasselbalch equation.(b) Explain, in terms of equilibrium, how this buffer resists a large pH change when a small amount of strong acid is added.

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(a) $\text{pH} = pK_a + \log\dfrac{[\text{A}^-]}{[\text{HA}]} = -\log(1.8\times10^{-5}) + \log\dfrac{0.25}{0.15} = 4.74 + 0.22 = 4.96$ (3 s.f., using unrounded $pK_a = 4.745$).(b) The buffer contains a large reserve of both the weak acid $\text{CH}_3\text{COOH}$ and its conjugate base $\text{CH}_3\text{COO}^-$. Added $\text{H}^+$ reacts with the conjugate base ($\text{CH}_3\text{COO}^- + \text{H}^+ \rightarrow \text{CH}_3\text{COOH}$), consuming most of the added acid and converting it to the weak acid form, so the equilibrium shifts to absorb the change and the $[\text{A}^-]/[\text{HA}]$ ratio, and hence the pH, changes only slightly.
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7

A $25.0\,\text{mL}$ sample of $0.10\,\text{M}$ acetic acid ($K_a = 1.8\times10^{-5}$) is titrated to its equivalence point with $0.10\,\text{M}$ $\text{NaOH}$, giving a total volume of $50.0\,\text{mL}$.(a) Explain why the pH at the equivalence point is greater than 7.(b) Given that the concentration of acetate ion at the equivalence point is $0.050\,\text{M}$, calculate the pH at the equivalence point ($K_b$ for $\text{CH}_3\text{COO}^- = \dfrac{K_w}{K_a}$).

Show answer
(a) At the equivalence point, all the acetic acid has been converted to its conjugate base, $\text{CH}_3\text{COO}^-$, which is a weak base and reacts with water to produce $\text{OH}^-$: $\text{CH}_3\text{COO}^- + \text{H}_2\text{O} \rightleftharpoons \text{CH}_3\text{COOH} + \text{OH}^-$. This makes the resulting solution basic, so pH $> 7$.(b) $K_b = \dfrac{1.0\times10^{-14}}{1.8\times10^{-5}} = 5.6\times10^{-10}$. Using $x^2/0.050 = K_b$: $x = [\text{OH}^-] = 5.27\times10^{-6}\,\text{M}$, $\text{pOH} = 5.28$, so $\text{pH} = 14.00 – 5.28 = 8.72$.
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8

For the reaction of ammonia with water, $\text{NH}_3\text{(aq)} + \text{H}_2\text{O(l)} \rightleftharpoons \text{NH}_4^{+}\text{(aq)} + \text{OH}^{-}\text{(aq)}$, identify the two Bronsted–Lowry conjugate acid-base pairs present, labeling each species as an acid or a base.

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Pair 1: $\text{H}_2\text{O}$ (acid, donates a proton) and $\text{OH}^-$ (its conjugate base). Pair 2: $\text{NH}_4^{+}$ (acid) and $\text{NH}_3$ (base, accepts a proton to form $\text{NH}_4^+$).
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Section C – Applications of thermodynamics (Unit 9)

Includes electrochemistry
9

For the Haber process, $\text{N}_2\text{(g)} + 3\text{H}_2\text{(g)} \rightarrow 2\text{NH}_3\text{(g)}$, $\Delta H^{\circ} = -92.2\,\text{kJ mol}^{-1}$ and $\Delta S^{\circ} = -198.7\,\text{J mol}^{-1}\text{K}^{-1}$.(a) Explain why $\Delta S^{\circ}$ is negative for this reaction.(b) Calculate $\Delta G^{\circ}$ at $298\,\text{K}$ and state whether the reaction is spontaneous under standard conditions at this temperature.(c) Calculate $\Delta G^{\circ}$ at $500\,\text{K}$ and explain why the sign of $\Delta G^{\circ}$ changes between the two temperatures.

Show answer
(a) The reaction converts 4 mol of gas (reactants) into 2 mol of gas (products), decreasing the number of gas particles and therefore the number of possible microstates, so entropy decreases.(b) $\Delta G^{\circ} = \Delta H^{\circ} – T\Delta S^{\circ} = -92.2 – (298)(-0.1987) = -92.2 + 59.2 = -33.0\,\text{kJ mol}^{-1}$. Since $\Delta G^{\circ} < 0$, the reaction is spontaneous at 298 K.(c) $\Delta G^{\circ} = -92.2 – (500)(-0.1987) = -92.2+99.35 = +7.15\,\text{kJ mol}^{-1}$. Since $\Delta H^{\circ}<0$ and $\Delta S^{\circ}<0$, the $-T\Delta S^{\circ}$ term is positive and grows with temperature; at high enough temperature it outweighs the negative $\Delta H^{\circ}$, making $\Delta G^{\circ}$ positive (non-spontaneous) at 500 K.
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10

A galvanic cell is constructed from the half-reactions $\text{Cu}^{2+}\text{(aq)} + 2e^- \rightarrow \text{Cu(s)}$, $E^{\circ} = +0.34\,\text{V}$, and $\text{Zn}^{2+}\text{(aq)} + 2e^- \rightarrow \text{Zn(s)}$, $E^{\circ} = -0.76\,\text{V}$.(a) Identify the cathode and anode half-reactions and calculate $E^{\circ}_{cell}$.(b) Calculate $\Delta G^{\circ}$ for the overall cell reaction, and use its sign to confirm the reaction is spontaneous as written.

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(a) The half-reaction with the more positive (less negative) reduction potential occurs as reduction at the cathode: $\text{Cu}^{2+} + 2e^- \rightarrow \text{Cu}$ (cathode). Zinc is oxidized at the anode: $\text{Zn} \rightarrow \text{Zn}^{2+} + 2e^-$. $E^{\circ}_{cell} = E^{\circ}_{cathode} – E^{\circ}_{anode} = 0.34 – (-0.76) = +1.10\,\text{V}$.(b) $n = 2$ mol electrons transferred. $\Delta G^{\circ} = -nFE^{\circ}_{cell} = -(2)(96{,}485)(1.10) = -2.12\times10^{5}\,\text{J mol}^{-1} = -212\,\text{kJ mol}^{-1}$. Since $\Delta G^{\circ} < 0$, the reaction is thermodynamically spontaneous under standard conditions, consistent with the positive $E^{\circ}_{cell}$.
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11

For the cell in the previous question ($\text{Zn}|\text{Zn}^{2+}\,||\,\text{Cu}^{2+}|\text{Cu}$, $E^{\circ}_{cell} = 1.10\,\text{V}$), the concentrations are changed to $[\text{Zn}^{2+}] = 1.0\,\text{M}$ and $[\text{Cu}^{2+}] = 0.010\,\text{M}$ at $298\,\text{K}$.(a) Write the reaction quotient, $Q$, for the overall cell reaction $\text{Zn(s)} + \text{Cu}^{2+}\text{(aq)} \rightarrow \text{Zn}^{2+}\text{(aq)} + \text{Cu(s)}$.(b) Use the Nernst equation, $E_{cell} = E^{\circ}_{cell} – \dfrac{RT}{nF}\ln Q$, to calculate $E_{cell}$ under these non-standard conditions.

Show answer
(a) $Q = \dfrac{[\text{Zn}^{2+}]}{[\text{Cu}^{2+}]} = \dfrac{1.0}{0.010} = 100$(b) $E_{cell} = 1.10 – \dfrac{(8.314)(298)}{(2)(96{,}485)}\ln(100) = 1.10 – 0.0592 = 1.04\,\text{V}$ (3 s.f.).
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12

For each process, predict the sign of $\Delta S^{\circ}_{sys}$ and justify your answer: (a) $\text{CaCO}_3\text{(s)} \rightarrow \text{CaO(s)} + \text{CO}_2\text{(g)}$, (b) $2\text{H}_2\text{(g)} + \text{O}_2\text{(g)} \rightarrow 2\text{H}_2\text{O(l)}$.

Show answer
(a) $\Delta S^{\circ}_{sys}$ is positive: a solid decomposes to give a solid and a gas, so the number of gas moles increases from 0 to 1, greatly increasing the number of accessible microstates.(b) $\Delta S^{\circ}_{sys}$ is negative: 3 mol of gas react to form 2 mol of liquid, so both the number of moles of gas decreases (to zero) and the particles end up in a much more ordered liquid state, decreasing entropy.
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