Answers to More Chapter 15 & 16 Study Questions
- Page ID
- 11240
\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)
\( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)
\( \newcommand{\dsum}{\displaystyle\sum\limits} \)
\( \newcommand{\dint}{\displaystyle\int\limits} \)
\( \newcommand{\dlim}{\displaystyle\lim\limits} \)
\( \newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\)
( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\)
\( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)
\( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\)
\( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)
\( \newcommand{\Span}{\mathrm{span}}\)
\( \newcommand{\id}{\mathrm{id}}\)
\( \newcommand{\Span}{\mathrm{span}}\)
\( \newcommand{\kernel}{\mathrm{null}\,}\)
\( \newcommand{\range}{\mathrm{range}\,}\)
\( \newcommand{\RealPart}{\mathrm{Re}}\)
\( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)
\( \newcommand{\Argument}{\mathrm{Arg}}\)
\( \newcommand{\norm}[1]{\| #1 \|}\)
\( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)
\( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\AA}{\unicode[.8,0]{x212B}}\)
\( \newcommand{\vectorA}[1]{\vec{#1}} % arrow\)
\( \newcommand{\vectorAt}[1]{\vec{\text{#1}}} % arrow\)
\( \newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)
\( \newcommand{\vectorC}[1]{\textbf{#1}} \)
\( \newcommand{\vectorD}[1]{\overrightarrow{#1}} \)
\( \newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}} \)
\( \newcommand{\vectE}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}} \)
\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)
\(\newcommand{\longvect}{\overrightarrow}\)
\( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)
\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)
- \(\mathrm{\dfrac{K_a}{[H^+]}=\dfrac{[C_3H_5O_2^-]}{[HC_3H_5O_2]}=\dfrac{1.3\times10^{-5}}{1.0\times10^{-5}}=1.3}\)
- \(\mathrm{pH=pK_a+\log\dfrac{[H_2BO_3^-]}{[H_3BO_3]}}\); \(\mathrm{pK_a=-\log(5.8\times10^{-10})=9.24}\); \(\mathrm{pH=9.24+\log\dfrac{1.5}{1}}\)
\(\mathrm{= 9.24 + 0.18 = 9.42}\)
- \(\mathrm{Na_2SO_4(aq) + Sr(NO_3)_2(aq) \rightarrow 2 NaNO_3(aq) + SrSO_4(s)}\)
\(\mathrm{SrSO_4(s) \rightleftharpoons Sr^{2+}(aq) + SO_4^{2-}(aq)}\) \(\mathrm{K_{sp} = [Sr^{2+}] \times [SO_4^{2-}] }\) \(\mathrm{K_{sp} = 3.2 \times 10^{-7}}\)
\(\mathrm{x = [Na_2SO_4] = [SO_4^{2-}]}\); \(\mathrm{[Sr^{2+}] = 0.10\: M}\)
\(\mathrm{3.2 \times 10^{-7} = (0.10\: M)(x)}\); \(\mathrm{x = \dfrac{3.2 \times 10^{-7}}{0.10\: M} = 3.2 \times 10^{-6}\: M}\)
- \(\mathrm{Pb(IO_3)_2(s) \rightleftharpoons Pb^{2+}(aq) + 2 IO_3^-(aq)}\); \(\mathrm{K_{sp} = [Pb^{2+}] \times [IO_3^-]^2}\)
\(\mathrm{[Pb^{2+}] = [Pb(IO_3)_2] = 2.6 \times 10^{-11}\: M}\); \(\mathrm{[IO_3^-] = [KIO_3] + [Pb(IO_3)_2] \approx [KIO_3] = 0.10\: M}\)
\(\mathrm{K_{sp} = [Pb^{2+}] \times [IO_3^-]^2 = (2.6 \times 10^{-11}\: M)(0.10\: M)^2 = 2.6 \times 10^{-13}}\)
- \(\mathrm{PbI_2(s) \rightleftharpoons Pb^{2+}(aq) + 2 I^-(aq)}\); \(\mathrm{K_{sp} = [Pb^{2+}] \times [I^-]^2}\); \(\mathrm{K_{sp} = 1.4 \times 10^{-8}}\)
\(\mathrm{x = [PbI_2] = [Pb^{2+}]}\); \(\mathrm{[I^-] = 0.010\: M + 2 [Pb^{2+}] \approx 0.010\: M}\)
\(\mathrm{K_{sp} = [Pb^{2+}] \times [I^-]^2}\); \(\mathrm{1.4 \times 10^{-8} = x [0.010\: M]^2}\); \(\mathrm{x = \dfrac{1.4 \times 10^{-8}}{1 \times 10^{-4}} = 1.4 \times 10^{-4}\: M}\)
- \(\mathrm{PbCl_2(s) \rightleftharpoons Pb^{2+}(aq) + 2 Cl^-(aq)}\); \(\mathrm{K_{sp} = [Pb^{2+}] \times [Cl^-]^2}\); \(\mathrm{K_{sp} = 1.7 \times 10^{-5}}\)
\(\mathrm{x = [PbCl_2] = [Pb^{2+}]}\); \(\mathrm{[Cl^-]= 2x}\); \(\mathrm{1.7 \times 10^{-5} = x (2x)^2 = 4 x^3}\)
\(\mathrm{x^3 = \dfrac{1.7 \times 10^{-5}}{4} = 4.2 \times 10^{-6}}\); \(\mathrm{x = (4.2 \times 10^{-6})^{1/3} = 0.016\: M}\)
- \(\mathrm{Pb(NO_3)_2(aq) + 2NaCl(aq) \rightarrow 2 NaNO_3(aq) + PbCl_2(s)}\)
\(\mathrm{PbCl_2(s) \rightleftharpoons Pb^{2+}(aq) + 2 Cl^-(aq)}\); \(\mathrm{K_{sp} = [Pb^{2+}] \times [Cl^-]^2}\); \(\mathrm{K_{sp} = 1.7 \times 10^{-5}}\)
\(\mathrm{x = [Pb(NO_3)_2] = [Pb^{2+}]}\); \(\mathrm{[Cl^-] = [NaCl] = 0.010\: M}\)
\(\mathrm{K_{sp} = [Pb^{2+}] \times [Cl^-]^2}\); \(\mathrm{1.7 \times 10^{-5} = x (0.010)^2}\); \(\mathrm{x = \dfrac{1.7 \times 10^{-5}}{1 \times 10^{-4}} = 0.17\: M}\)
-
- 0.10 M \(\ce{NaOH}\); \(\mathrm{pH = 13 \rightarrow yellow}\)
- 0.10 M \(\ce{NaOH}\); \(\mathrm{pH = 13 \rightarrow purple}\)
- cresol red is orange when \(\mathrm{pH = pKa}\); \(\mathrm{pH = 1}\)
- yellow in methyl yellow: \(\mathrm{pH > 4}\); yellow in cresol purple: \(\mathrm{pH < 7}\); so, \(\mathrm{4 < pH < 7}\)
-
- \(\mathrm{moles\: base = moles\: acid = 28.0\: mL \times\dfrac{0.150\:moles\:HCl}{1000\:mL} = 0.00420\: moles}\)
- \(\mathrm{molar\: mass =\dfrac{mass}{moles}=\dfrac{0.290\:g}{0.00420\:moles} = 69.0\: g/mole}\)
-
- \(\mathrm{\textrm{moles base = moles acid: } V_B \times M_B = V_A \times M_A}\)
\(\mathrm{2.50\: mL \times 3.00\: M = 750\: mL \times M_A}\); \(\mathrm{M_A = \dfrac{2.50\times3.00}{750}=0.0100\: M\:HCl}\)
- \(\mathrm{pH = -\log[H^+] = -\log (1.00 \times 10^{-2}\: M)}\); \(\mathrm{pH = 2.0}\)