# Answers to Chapter 15 & 16 Study Questions

1. strong acid + strong base:  steepest at equivalence, low pH at start depending on concentration of acid, equivalence at pH 7, no buffering.
2. weak acid + strong base:  less steep at equivalence than strong acid, low pH at start depending on $$\mathrm{K_a}$$ and concentration of acid, equivalence at basic pH, buffering.
3. weak base + strong acid:  less steep at equivalence than strong base, high pH at start depending on $$\mathrm{K_b}$$ and concentration of base, equivalence at acid pH, buffering.

1. blue
2. $$\mathrm{pH = 5}$$
3. an indicator changes color when $$\mathrm{[HA] = [A^-]}$$ and $$\mathrm{[H^+] = K_a}$$;
therefore $$\mathrm{K_a = [H^+] = 10^{-3.5} = 3.2 \times 10^{-4}}$$
4. weak base + strong acid $$\ce{\rightarrow}$$ weak acid;   acid endpoint;  methyl red

1. $$\mathrm{V_A \times M_A = V_B \times M_B}$$;  $$\mathrm{V_A \times 2.00\: M = 12.5\: mL \times 0.800 \:M}$$; $$\mathrm{V_A = 0.500\: mL}$$

(optional pH of solution:)  $$\mathrm{H^+ + NH_3 \rightarrow NH_4^+}$$; $$\mathrm{final\: volume = 12.5 + 0.5 = 13.0\: mL}$$

Final $$\mathrm{[NH_4^+]}$$:  $$\mathrm{12.5\: mL \times 0.0800\: M = 13.0\: mL \times M_2}$$;  $$\mathrm{M_2 = 0.0769\: M\: NH_4^+}$$ at end.

So, calculate the pH of 0.0769 M $$\mathrm{NH_4^+}$$:

$$\mathrm{K_a =\dfrac{[H^+]\times[NH_3]}{[NH_4^+]}}$$; $$\mathrm{5.6\times10^{-10}=\dfrac{x^2}{0.769\:M}}$$;  $$\mathrm{x^2 = (5.6 \times 10^{-10})(0.0769) = 4.3 \times 10^{-11}}$$;

$$\mathrm{x = (4.3 \times 10^{-11})^{1/2} = 6.6 \times 10^{-6}\: M}$$;  $$\mathrm{pH = - \log (6.6 \times 10^{-6}\: M) = 5.2}$$

(you get the same answer if you used 0.080 M $$\mathrm{NH_4^+}$$.)

1. $$\mathrm{K_a(CH_3COOH) = 1.8 \times 10^{-5}}$$;  $$\mathrm{[H^+] = 1.0 \times 10^{-5}\: M}$$
$$\mathrm{\dfrac{[CH_3COO^-]}{[CH_3COOH]}=\dfrac{K_a}{[H^+]}=\dfrac{1.8\times10^{-5}}{1.0\times10^{-5}}=1.8}$$
2. Mix the solutions in a 1.8: 1 ratio.  For example, mix 180. mL of 0.100 M $$\mathrm{NaCH_3COO}$$ with 100 mL 0.100 M $$\mathrm{CH_3COOH}$$.
3. Mix 280 mL 0.100 M $$\mathrm{CH_3COOH}$$ and 180. mL 0.100 M $$\ce{NaOH}$$.
4. Another pH 5.00 buffer might be $$\ce{Al(H2O)6^3+}$$ or even benzoic acid.

1. limiting reactant problem:  $$\mathrm{0.750\: L \times 0.400\: M\: NaOH = 0.300 \:moles\: OH^-}$$
$$\mathrm{0.250\: L \times 0.800\: M\: HCl = 0.200\: mol\: H^+}$$.
$$\mathrm{0.300\: moles\: OH^- + 0.200\: mol\: H^+ \rightarrow 0.200\: mol\: H_2O + 0.100\: mol\: OH^-}$$ remain.
$$\mathrm{Total\: volume = 250\: mL + 750\: mL = 1.00\: liter}$$.
$$\mathrm{[OH^-]=\dfrac{0.100\:mol}{1\:liter}=1.00\:M}$$; $$\mathrm{pOH = 1.0}$$; $$\mathrm{pH = 13.0}$$

2. $$\mathrm{pH=pK_a+\log\dfrac{[HCO_3^-]}{[H_2CO_3]}}$$;     $$\mathrm{pK_a=-\log(4.4\times10^{-7})=6.36}$$
$$\mathrm{pH=6.36+\log\left( \dfrac{1}{2} \right)= 6.36 - 0.30 = 6.06}$$

1. $$\mathrm{MgCl_­2 \rightleftarrows Mg^{2+}(aq) + 2 Cl^-(aq)}$$;    $$\mathrm{K_{sp} = [Mg^{2+}] \times [Cl^-]^2}$$

$$\mathrm{[MgCl_2]=\dfrac{8.0\:g\:MgCl_2}{108\:g\:solution}\times\dfrac{1\:g\:solution}{1\:mL\:solution}\times\dfrac{1000\:mL}{1\:L}\times\dfrac{1\:mol\:MgCl_2}{95.2\:g\:MgCl_2}=0.778\:M}$$

0.778 M $$\ce{MgCl­2}$$:  $$\mathrm{[Mg^{2+}] = 0.778\: M}$$;  $$\mathrm{[Cl^-] = 2(0.778\: M) = 1.56\: M}$$

$$\mathrm{K_{sp} = [Mg^{2+}] \times [Cl^-]^2 = (0.778) \times (1.56)^2 = 1.88}$$

1. $$\mathrm{CuCrO_­4 \rightleftarrows Cu^{2+}(aq) + CrO_4^{2-}(aq)}$$;    $$\mathrm{K_{sp} = [Cu^{2+}] \times [CrO_4^{2-}] = 3.6 \times 10^{-6}}$$;

$$\mathrm{x = [CuCrO­_4] = [Cu^{2+}] = [CrO_4^{2-}]}$$;  $$\mathrm{K_{sp} = 3.6 \times 10^{-6} = x^2}$$

$$\mathrm{x = (3.6 \times 10^{-6})^{1/2} = 1.9 \times 10^{-3}\: M}$$

1. $$\mathrm{2 AgNO_3(aq) + Na_2CO_3(aq) \rightarrow 2 NaNO_3(aq) + Ag_2CO_3(s)}$$

$$\mathrm{Ag_2CO_3(s) \rightleftarrows 2 Ag^+(aq) + CO_3^{2-}(aq)}$$;  $$\mathrm{K_{sp} = [Ag^+]^2 \times [CO_3^{2-}] = 8.1 \times 10^{-12}}$$

$$\mathrm{x = [Ag^+]}$$;  $$\mathrm{[CO_3^{2-}] = 0.020\: M}$$;  $$\mathrm{K_{sp} = x^2(0.020) = 8.1 \times 10^{-12}}$$

$$\mathrm{x^2=\dfrac{8.1\times10^{-12}}{0.020}= 4.1 \times 10^{-10}}$$;  $$\mathrm{x = (4.1 \times 10^{-10})^{1/2} = 2.0 \times 10^{-5}\: M}$$

1. $$\mathrm{Pb(NO_3)_2(aq) + 2 NaBr(aq) \rightarrow 2 NaNO_3(aq) + PbBr_2(s)}$$.

$$\mathrm{PbBr_2(s) \rightleftarrows Pb^{2+}(aq) + 2 Br^-(aq)}$$;    $$\mathrm{K_{sp} = [Pb^{2+}] \times [Br^-]^2}$$;   $$\mathrm{K_{sp} = 4.6 \times 10^{-6}}$$

When solutions are mixed in a 1:1 ratio, each is diluted by a factor of 2.

$$\mathrm{[Pb^{2+}] = \dfrac{0.0100\: M}{2} = 0.00500 \:M}$$;  $$\mathrm{[Br^-] = \dfrac{0.0200\: M}{2} = 0.0100\: M}$$

$$\mathrm{Q = [Pb^{2+}] \times [Br^-]^2 = (0.00500) \times (0.0100)^2 = 5.0 \times 10^{-7}}$$

$$\mathrm{5.0 \times 10^{-7} < 4.6 \times 10^{-6}}$$;  $$\mathrm{Q < K_{sp}}$$;  therefore, no precipitate forms

1. strong acid reacts with the weak acid in the buffer:

$$\mathrm{H^+(aq) + HCO_3^-(aq) \rightarrow H_2CO_3(aq)}$$

1. $$\mathrm{Ag_3PO_4(s) \rightleftarrows 3 Ag^+(aq) + PO_4^{3-}(aq)}$$;  $$\mathrm{K_{sp} = [Ag^+]^3 \times [PO_4^{2-}]}$$

1. $$\mathrm{molar\: mass = \dfrac{mass}{moles}}$$;

$$\textrm{moles acid = moles base: }\mathrm{17.5\:mL\times\dfrac{0.268\:mol\:NaOH}{1000\:mL} = 0.00469\: moles\: acid}$$

$$\mathrm{molar\:mass = \dfrac{0.422\:g}{0.00469\:moles}= 90.0\: g/mol}$$