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# Answers to More Chapter 15 & 16 Study Questions

1. $$\mathrm{\dfrac{K_a}{[H^+]}=\dfrac{[C_3H_5O_2^-]}{[HC_3H_5O_2]}=\dfrac{1.3\times10^{-5}}{1.0\times10^{-5}}=1.3}$$

1. $$\mathrm{pH=pK_a+\log\dfrac{[H_2BO_3^-]}{[H_3BO_3]}}$$; $$\mathrm{pK_a=-\log(5.8\times10^{-10})=9.24}$$; $$\mathrm{pH=9.24+\log\dfrac{1.5}{1}}$$

$$\mathrm{= 9.24 + 0.18 = 9.42}$$

1. $$\mathrm{Na_2SO_4(aq) + Sr(NO_3)_2(aq) \rightarrow 2 NaNO_3(aq) + SrSO_4(s)}$$

$$\mathrm{SrSO_4(s) \rightleftharpoons Sr^{2+}(aq) + SO_4^{2-}(aq)}$$ $$\mathrm{K_{sp} = [Sr^{2+}] \times [SO_4^{2-}] }$$ $$\mathrm{K_{sp} = 3.2 \times 10^{-7}}$$

$$\mathrm{x = [Na_2SO_4] = [SO_4^{2-}]}$$; $$\mathrm{[Sr^{2+}] = 0.10\: M}$$

$$\mathrm{3.2 \times 10^{-7} = (0.10\: M)(x)}$$; $$\mathrm{x = \dfrac{3.2 \times 10^{-7}}{0.10\: M} = 3.2 \times 10^{-6}\: M}$$

1. $$\mathrm{Pb(IO_3)_2(s) \rightleftharpoons Pb^{2+}(aq) + 2 IO_3^-(aq)}$$; $$\mathrm{K_{sp} = [Pb^{2+}] \times [IO_3^-]^2}$$

$$\mathrm{[Pb^{2+}] = [Pb(IO_3)_2] = 2.6 \times 10^{-11}\: M}$$; $$\mathrm{[IO_3^-] = [KIO_3] + [Pb(IO_3)_2] \approx [KIO_3] = 0.10\: M}$$

$$\mathrm{K_{sp} = [Pb^{2+}] \times [IO_3^-]^2 = (2.6 \times 10^{-11}\: M)(0.10\: M)^2 = 2.6 \times 10^{-13}}$$

1. $$\mathrm{PbI_2(s) \rightleftharpoons Pb^{2+}(aq) + 2 I^-(aq)}$$; $$\mathrm{K_{sp} = [Pb^{2+}] \times [I^-]^2}$$; $$\mathrm{K_{sp} = 1.4 \times 10^{-8}}$$

$$\mathrm{x = [PbI_2] = [Pb^{2+}]}$$; $$\mathrm{[I^-] = 0.010\: M + 2 [Pb^{2+}] \approx 0.010\: M}$$

$$\mathrm{K_{sp} = [Pb^{2+}] \times [I^-]^2}$$; $$\mathrm{1.4 \times 10^{-8} = x [0.010\: M]^2}$$; $$\mathrm{x = \dfrac{1.4 \times 10^{-8}}{1 \times 10^{-4}} = 1.4 \times 10^{-4}\: M}$$

1. $$\mathrm{PbCl_2(s) \rightleftharpoons Pb^{2+}(aq) + 2 Cl^-(aq)}$$; $$\mathrm{K_{sp} = [Pb^{2+}] \times [Cl^-]^2}$$; $$\mathrm{K_{sp} = 1.7 \times 10^{-5}}$$

$$\mathrm{x = [PbCl_2] = [Pb^{2+}]}$$; $$\mathrm{[Cl^-]= 2x}$$; $$\mathrm{1.7 \times 10^{-5} = x (2x)^2 = 4 x^3}$$

$$\mathrm{x^3 = \dfrac{1.7 \times 10^{-5}}{4} = 4.2 \times 10^{-6}}$$; $$\mathrm{x = (4.2 \times 10^{-6})^{1/3} = 0.016\: M}$$

1. $$\mathrm{Pb(NO_3)_2(aq) + 2NaCl(aq) \rightarrow 2 NaNO_3(aq) + PbCl_2(s)}$$

$$\mathrm{PbCl_2(s) \rightleftharpoons Pb^{2+}(aq) + 2 Cl^-(aq)}$$; $$\mathrm{K_{sp} = [Pb^{2+}] \times [Cl^-]^2}$$; $$\mathrm{K_{sp} = 1.7 \times 10^{-5}}$$

$$\mathrm{x = [Pb(NO_3)_2] = [Pb^{2+}]}$$; $$\mathrm{[Cl^-] = [NaCl] = 0.010\: M}$$

$$\mathrm{K_{sp} = [Pb^{2+}] \times [Cl^-]^2}$$; $$\mathrm{1.7 \times 10^{-5} = x (0.010)^2}$$; $$\mathrm{x = \dfrac{1.7 \times 10^{-5}}{1 \times 10^{-4}} = 0.17\: M}$$

1. 0.10 M $$\ce{NaOH}$$; $$\mathrm{pH = 13 \rightarrow yellow}$$
2. 0.10 M $$\ce{NaOH}$$; $$\mathrm{pH = 13 \rightarrow purple}$$
3. cresol red is orange when $$\mathrm{pH = pKa}$$; $$\mathrm{pH = 1}$$
4. yellow in methyl yellow: $$\mathrm{pH > 4}$$; yellow in cresol purple: $$\mathrm{pH < 7}$$; so, $$\mathrm{4 < pH < 7}$$

1. $$\mathrm{moles\: base = moles\: acid = 28.0\: mL \times\dfrac{0.150\:moles\:HCl}{1000\:mL} = 0.00420\: moles}$$
2. $$\mathrm{molar\: mass =\dfrac{mass}{moles}=\dfrac{0.290\:g}{0.00420\:moles} = 69.0\: g/mole}$$

1. $$\mathrm{\textrm{moles base = moles acid: } V_B \times M_B = V_A \times M_A}$$

$$\mathrm{2.50\: mL \times 3.00\: M = 750\: mL \times M_A}$$; $$\mathrm{M_A = \dfrac{2.50\times3.00}{750}=0.0100\: M\:HCl}$$

1. $$\mathrm{pH = -\log[H^+] = -\log (1.00 \times 10^{-2}\: M)}$$; $$\mathrm{pH = 2.0}$$