Answers to More Chapter 15 & 16 Study Questions

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\(\mathrm{\dfrac{K_a}{[H^+]}=\dfrac{[C_3H_5O_2^-]}{[HC_3H_5O_2]}=\dfrac{1.3\times10^{-5}}{1.0\times10^{-5}}=1.3}\)

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

\(\mathrm{= 9.24 + 0.18 = 9.42}\)

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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