Calculate the concentration of all solute species in each of the following solutions of acids or bases. Assume that the ionization of water can be neglected, and show that the change in the initial concentrations can be neglected. Ionization constants can be found in Appendix H and Appendix I.

\((a) 0.0092M HClO\), a weak acid

\((b) 0.0784M {C_6}{H_5}N{H_2}\), a weak base

\((c) 0.0810{\rm{ }}M HCN\), a weak acid

\((d) 0.11M{\left( {C{H_3}} \right)_3}\;N\), a weak base

\((e) 0.120MFe\left( {{H_2}O} \right)_6^{2 + }\), a weak acid \({K_a} = 1.6 \times 1{0^{ - 7}}\)

\(\begin{aligned}{\bf{HClO(aq)}} \to {{\bf{H}}^{\bf{ + }}}{\bf{(aq) + }}{{\bf{l}}^{\bf{ - }}}{\bf{(aq)}}\\{{\bf{K}}_{\bf{a}}}{\bf{ = }}\;\frac{{\left( {{{\bf{H}}^{\bf{ + }}}} \right){\bf{ \times }}\left( {{\bf{Cl}}{{\bf{O}}^{\bf{ - }}}} \right)}}{{\left( {{\bf{HClO}}} \right)}}.\end{aligned}\)

We can designate\({{\bf{H}}^{\bf{ + }}}\)and\({\bf{Cl}}{{\bf{O}}^{\bf{ - }}}\)as x and presume that their concentrations are the same. We have 0.0092 M of HCl at the start of the reaction, as well as 0 M of\({{\bf{H}}^{\bf{ + }}}\)and\({\bf{Cl}}{{\bf{O}}^{\bf{ - }}}\). When the reaction begins, the concentration of HClO falls by x, while the concentrations of\({{\bf{H}}^{\bf{ + }}}\)and\({\bf{Cl}}{{\bf{O}}^{\bf{ - }}}\)ions grow by x, until the equilibrium is attained. As a result, the equilibrium concentrations are as follows:

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\(\begin{aligned}\left( {HClO} \right) = 0.0092 - x\\\;\;\;\;\left( {{H^ + }} \right) = \left( {Cl{O^ - }} \right) = x.\end{aligned}\)

Now the equation for \({K_a}\) will look like this:

\({K_a} = \frac{{{x^2}}}{{0.0092 - x}}.\)

We can assume that \(0.0092 - x\) is almost equivalent to \(0.0092\) because the \({K_a}\) value is so low.

\(\begin{aligned}\;\;\;\;\;\;\;\;\;{K_a} = \frac{{{x^2}}}{{0.0092}}\\2.9 \times 1{0^{ - 8}} = \frac{{{x^2}}}{{0.0092}}\\\;\;\;\;\;\;\;\;\;\;\;\;x = 1.63 \times 1{0^{ - 5}}.\end{aligned}\)

As a result, we may be confident that the change in concentration for the acid molecule can be ignored.

\(\frac{{1.63 \times 1{0^{ - 5}}}}{{0.0092}} = 0.002 = 0.2\% .\)

Because it is less than \(1\) percent of the initial value, \({H^ + }\)and \(Cl{O^ - }\)ions have a concentration of \(1.63 \times 1{0^{ - 5}}\;M,\) whereas has a concentration of \(Cl{O^ - }\)is \(0.0092\;M.\)

b. The reaction in question is:

\({C_6}{H_5}N{H_2}(aq) + {H_2}O(l) \to {C_6}{H_5}NH_3^ + (aq) + O{H^ - }(aq).\)

The \({K_a}\)value is:

\({K_a} = \frac{{\left( {{C_6}{H_5}NH_3^ + } \right) \times \left( {O{H^ - }} \right)}}{{\left( {{C_6}{H_5}N{H_2}} \right)}}.\)

Again, we assume that the concentrations of \({C_6}{H_5}NH_3^ + \) and \(O{H^ - }\)are the same, and that the change in \({C_6}{H_5}N{H_2}\) is negligible:

\(\begin{aligned}\;\;\;\;\;\;\;\;\;\;\;{K_a} = \frac{{{x^2}}}{{0.0784}}\\4.3 \times 1{0^{ - 10}} = \frac{{{x^2}}}{{0.0784}}\\\;\;\;\;\;\;\;\;\;\;\;\;\;\;x = 5.8 \times 1{0^{ - 6}}\;M.\end{aligned}\)

The assumption that the concentration change of \({C_6}{H_5}N{H_2}\) is small, is accurate because:

\(\frac{{5.8 \times 1{0^{ - 6}}}}{{0.0784}} = 7.4 \times 1{0^{ - 5}}.\)

Less than \(1\) percent change in the concentration of \(O{H^ - }\) and \({C_6}{H_5}NH_3^ + \) is \(5.8 \times 1{0^{ - 6}}\,M,\) and the concentration of \({C_6}{H_5}N{H_2}\) is \(0.0784\;M.\)

\(HCN(aq) \to {H^ + }(aq) + C{N^ - }(aq).\)

The \({K_a}\) value is:

\({K_a} = \frac{{\left( {{H^ + }} \right) \times \left( {C{N^ - }} \right)}}{{\left( {HCN} \right)}}.\)

We may assume that the concentrations of \({H^ + }\) and \(C{N^ - }\)are the same and that the concentration change for \(HCN\) is insignificant once again.

\(\begin{aligned}4.9 \times 1{0^{ - 10}} = \frac{{{x^2}}}{{0.0810}}\\\;\;\;\;\;\;\;\;\;\;\;\;\;\;x = 6.3 \times 1{0^{ - 6}} M.\end{aligned}\)

The assumption that the decrease in \(HCN\) concentration is negligible is true because:

\(\frac{{6.3 \times 1{0^{ - 6}}}}{{0.0810}} = 7.8 \times 1{0^{ - 5}}.\)

Less than \(1\) percent change in the concentration of \({H^ + }\) and \(C{N^ - }\) is \(6.3 \times 1{0^{ - 6}} M,\) and the concentration of \(HCN\) is \(0.0810 M.\)

d. The reaction in question is:

\(\begin{aligned}{\left( {C{H_3}} \right)_3}\;N(aq) + {H_2}O(l) \to {\left( {C{H_3}} \right)_3}N{H^ + }(aq) + O{H^ - }(aq).\\\\The {K_a} value is:\\{K_a} = \frac{{\left( {{{\left( {C{H_3}} \right)}_3}N{H^ + }} \right) \times \left( {O{H^ - }} \right)}}{{\left( {{{\left( {C{H_3}} \right)}_3}N} \right)}}.\end{aligned}\)

We can assume that the concentrations of \({\left( {C{H_3}} \right)_3}N{H^ + }\) and \(O{H^ - }\) are the same, and that the change in the concentration of \({\left( {C{H_3}} \right)_3}N\) is negligible.

\(\begin{aligned}6.3 \times 1{0^{ - 5}} = \frac{{{x^2}}}{{0.11}}\\\;\;\;\;\;\;\;\;\;\;\;\;x = 2.6 \times 1{0^{ - 3}}\;M.\end{aligned}\)

The assumption that the change in \({\left( {C{H_3}} \right)_3}N\) concentration is negligible is valid because:

\(\frac{{2.6 \times 1{0^{ - 3}}}}{{0.11}} = 0.02.\)

Less than \(2\) percent change in the concentration of \({\left( {C{H_3}} \right)_3}N{H^ + }\) and \(O{H^ - }\) is \(2.6 \times 1{0^{ - 3}} M,\) and the concentration of \({\left( {C{H_3}} \right)_3}N\) is \(0.11 M.\)

\(\begin{aligned}{\left( {Fe{{\left( {{H_2}O} \right)}_6}} \right)^{2 + }}(aq) \to {\left( {Fe{{\left( {{H_2}O} \right)}_5}OH} \right)^ + }(aq) + {H^ + }(aq).\\\\The {K_a} value is:\\{K_a} = \frac{{{{\left( {Fe{{\left( {{H_2}O} \right)}_5}OH} \right)}^ + } \times \left( {{H^ + }} \right)}}{{{{\left( {Fe{{\left( {{H_2}O} \right)}_6}} \right)}^{2 + }}}}.\end{aligned}\)

We can assume that the concentrations of \({\left( {Fe{{\left( {{H_2}O} \right)}_5}OH} \right)^ + }\) and \({H^ + }\) are the same, and that the concentration change of \({\left( {Fe{{\left( {{H_2}O} \right)}_6}} \right)^{2 + }}\) is inconsequential once again.

\(\begin{aligned}1.6 \times 1{0^{ - 7}} = \frac{{{x^2}}}{{0.120}}\\\;\;\;\;\;\;\;\;\;\;\;\;x = 1.4 \times 1{0^{ - 4}} M.\end{aligned}\)

The assumption that the change in \({\left( {Fe{{\left( {{H_2}O} \right)}_5}OH} \right)^ + }\) concentration is negligible is accurate because:

\(\frac{{1.4 \times 1{0^{ - 4}}}}{{0.120}} = 0.001.\)

Less than \(1\) percent change in the concentration of \({\left( {Fe{{\left( {{H_2}O} \right)}_5}OH} \right)^ + }\) and \({H^ + }\) is \(1.4 \times 1{0^{ - 4}} M,\) and the concentration of \({\left( {Fe{{\left( {{H_2}O} \right)}_6}} \right)^{2 + }}\) is. \(0.120 M.\)