Question

Why can we ignore the contribution of water to the concentrations of\({H_3}{O^ + }\)in the solutions of following acids:\(0.0092MHClO\), a weak acid\(0.0810MHCN\), a weak acid but not the contribution of water to the concentration of\(O{H^ - }\)?

Short answer

\(HClO\)- We can see that the \(O{H^ - }\)concentration in the solution is almost a thousand times smaller than the water concentration and we cannot thus ignore the \(O{H^ - }\) concentration in water in our calculations.

\(HCN\)- We can see that the \(O{H^ - }\)concentration in the solution is a hundred times smaller than the water concentration and we cannot thus ignore the \(O{H^ - }\)concentration in water in our calculations.

\({\left( {Fe{{\left( {{H_2}O} \right)}_6}} \right)^{2 + }}\) - We can see that the \({\rm{O}}{{\rm{H}}^ - }\) concentration in the solution is many times smaller than the water concentration and we cannot thus ignore the \(O{H^ - }\) concentration in water in our calculations Step-by-step-solution.

Step-by-step solution

Step 1:

Generally speaking, we can ignore a certain value if it is a 100 or \(1{0^2}\) times smaller than the value in question. In this case, we need to calculate the concentration of \({H_3}{O^ + }\)and \(O{H^ - }\)ions in the solution. If the value is \(1{0^2}\) or 10 times greater than that from water \(\left( {1{0^{ - 7}}M} \right),\)we can ignore it, if it is not, we have to consider it in our calculations. The concentrations of \({H_3}{O^ + }\)and \(O{H^ - }\)ions in pure water are both\(1{0^{ - 7}}M\), which is a constant In order to calculate hthe \({H_3}{O^ + }\)and \(O{H^ - }\) concentrations of the acid solutions in the problem, we have to use their \({K_a}\) values. We can look up the \({K_a}\) values for each acid in the table in Appendix \(H\) in the book.

To find the OH- concentration in M HCLO:

For \(HClO,\) the \({K_a}\) value is \(2.9 \times 1{0^{ - 8}}\). The ionization reaction for \(HClO\) looks like this:

\(HClO(aq) \to Cl{O^ - }(aq) + {H^ + }(aq)\)

The \({K_a}\) value is calculated as:

\({K_a} = \frac{{c{{\left( {Cl{O^ - }} \right)}_{eq}} \times c{{\left( {{H^ + }} \right)}_{eq}}}}{{c{{(HClO)}_{eq}}}}\)

Same as the initial concentration because, given the very small \({K_a}\)value, the concentration change of \(HClO\)is very small. Now the \({K_a}\) equation looks

\(\begin{ALIGN}{*{20}{2.9 \times {{10}^{ - 8}} &= \frac{{{x^2}}}{{0.0092}}}\\{{x^2} &= 2.7 \times {{10}^{ - 10}}}\\{x &= c\left( {{H^ + }} \right) &= 1.6 \times {{10}^{ - 5}}M}\end{ALIGN}\)

We can see that the \({H_3}{O^ + }\) concentration is about 100 times greater than the water concentration and we can therefore ignore it. We can calculate \(O{H^ - }\) concentration from the ionic water product:

\(\begin{align}{{20}}{{K_w} = {{10}^{ - 14}}}\\{{K_w} = c\left( {{H^ + }} \right) \times c\left( {O{H^ - }} \right)}\\{c\left( {O{H^ - }} \right) = \frac{{{K_w}}}{{c\left( {{H^ + }} \right)}}}\\{c\left( {O{H^ - }} \right) = \frac{{{{10}^{ - 14}}}}{{1.6 \times {{10}^{ - 5}}}}}\\{c\left( {O{H^ - }} \right) = 6.25 \times {{10}^{ - 10}}M}\end{align}\)

We can see that the \(O{H^ - }\) concentration in the solution is almost a thousand times smaller than the water concentration and we cannot thus ignore the \(OH\) concentration in water in our calculations.

To find the OH- concentration in 0.0810 M HCN:

For \(HCN,\) \({K_a}\) value is\(4.9 \times 1{0^{ - 10}}\).The ionization reaction:

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

We can assume that the concentrations of\(C{N^ - }\)and\({H^ + }\)are the same and mark them as\(x\). We can also assume that the equilibrium concentration of HCN is the same as the initial concentration because, given the very small\({K_a}\)value, the concentration change of\(HCN\)is very small. Now the\({K_a}\)equation looks like this:

\(\begin{align}{{20}}{4.9 \times {{10}^{ - 10}} = \frac{{{x^2}}}{{0.0810}}}\\{{x^2} = 4.0 \times {{10}^{ - 11}}}\\{x = c\left( {{H^ + }} \right) = 6.3 \times {{10}^{ - 6}}M}\end{align}\)

We can see that the\({H_3}{O^ + }\)concentration is much greater than the water concentration and we can therefore ignore it. We can calculate the\(O{H^ - }\)concentration from the ionic water product:

\(\begin{align}{{20}}{{K_w} = {{10}^{ - 14}}}\\{{K_w} = c\left( {{H^ + }} \right) \times c\left( {O{H^ - }} \right)}\\{c\left( {O{H^ - }} \right) = \frac{{{{10}^{ - 14}}}}{{6.3 \times {{10}^{ - 6}}}}}\\{c\left( {O{H^ - }} \right) = 1 \times {{10}^{ - 9}}M}\end{align}\)

We can see that the \(O{H^ - }\) concentration in the solution is a hundred times smaller than the water concentration and we cannot thus ignore \(OH\) - concentration in water in our calculations.

To find the OH- concentration in 0.120 M (Fe (H2o)6)2+:

For\({\left( {Fe{{\left( {{H_2}O} \right)}_6}} \right)^{2 + }}\), the\({K_a}valueis\)1.6\(1{0^{ - 7}}\). The ionization reaction:

\({\left( {{\rm{Fe}}{{\left( {{{\rm{H}}_2}{\rm{O}}} \right)}_6}} \right)^{2 + }}(aq) + {{\rm{H}}_2}{\rm{O}}({\rm{l}}) \to {\left( {{\rm{Fe}}{{\left( {{{\rm{H}}_2}{\rm{O}}} \right)}_6}{\rm{OH}}} \right)^ + }(aq) + {{\rm{H}}^ + }(aq)\)

We can assume that the concentrations\(of{\left( {Fe{{\left( {{H_2}O} \right)}_6}OH} \right)^ + }\)and\({H^ + }\)are the same and mark them as\(x\). We can also assume that the equilbrium concentration of\({\left( {Fe{{\left( {{H_2}O} \right)}_6}} \right)^{2 + }}\)is the same as the initial concentration because, given the very small\({K_a}\)value, the concentration change\(of {\left( {Fe{{\left( {{H_2}O} \right)}_6}} \right)^{2 + }}\)is very small. Now the\({K_a}\)equation looks like this:

\(\begin{align}{{20}}{1.6 \times {{10}^{ - 7}} = \frac{{{x^2}}}{{0.120}}}\\{{x^2} = 1.9 \times {{10}^{ - 8}}}\\{x = c\left( {{H^ + }} \right) = 1.4 \times {{10}^{ - 4}}M}\end{align}\)

We can see that the \({H_3}{O^ + }\) concentration is more than a 1000 times greater than the water concentration and we can therefore ignore it. We can calculate \(O{H^ - }\) concentration from the ionic water product:

\(\begin{align}{{20}{c}}{{K_w} = {{10}^{ - 14}}}\\{{K_w} = c\left( {{H^ + }} \right) \times c\left( {O{H^ - }} \right)}\\{c\left( {O{H^ - }} \right) = \frac{{{K_w}}}{{c\left( {{H^ + }} \right)}}}\\{c\left( {O{H^ - }} \right) = \frac{{{{10}^{ - 14}}}}{{1.4 \times {{10}^{ - 4}}}}}\\{c\left( {O{H^ - }} \right) = 7.1 \times {{10}^{-11}}M}\end{align}\).

We can see that the \(O{H^ - }\) concentration in the solution is many times smaller than the water concentration and we cannot thus ignore the \(O{H^ - }\) concentration in water in our calculations.