The \(pH\) of a solution of household ammonia, a \(0.950M\) solution of \(N{H_3}\), is \(11.612.\) Determine \({K_b} \)for \(N{H_3}\) from these data.

The following is the response in question:

\({\bf{N}}{{\bf{H}}_{\bf{3}}}{\bf{(aq) + }}{{\bf{H}}_{\bf{2}}}{\bf{O(l)}} \to {\bf{NH}}_{\bf{4}}^{\bf{ + }}{\bf{(aq) + O}}{{\bf{H}}^{\bf{ - }}}{\bf{(aq)}}.\)

We need to know the concentrations of OH-,\({\bf{NH}}_{\bf{4}}^{\bf{ + }}\), and\({\bf{N}}{{\bf{H}}_{\bf{3}}}\)at equilibrium in order to compute the\({{\bf{K}}_{\bf{b}}}\)value.

From the \(pH\), we can compute the concentration of \(O{H^ - }:\)

\(\begin{aligned}pOH = 14 - pH\\\;\;\;\;\;\;\; = 14 - 11.612\\\;\;\;\;\;\;\; = 2.388.\end{aligned}\)

\(\begin{aligned}\;\;\;pOH &= - log\left( {O{H^ - }} \right)\\\left( {O{H^ - }} \right) &= 1{0^{ - pOH}} M.\\\; &= 1{0^{ - 2.388}} M\\\; &= 4.093 \times 1{0^{ - 3}} M.\end{aligned}\)

We can assume that the concentrations of \(O{H^ - }\) and \(NH_4^ + \) are the same. The starting concentration of \(N{H_3}\) is \(0.950 M\) text, while both the initial concentrations of \(NH_4^ + \) and \(O{H^ - }\) are \(0 M\). The concentration of \(N{H_3}\) lowers by \(x\)while the concentration of the ions rises by \(x\) until the equilibrium is established. We can determine the equilibrium concentration of \(N{H_3}\) because we know the equilibrium concentrations of the ions and the value of \(x\):

\(\begin{aligned}\left( {N{H_3}} \right) &= 0.950 M - x\\\; &= \left( {0.950 - 4.093 \times 1{0^{ - 3}}} \right) M\\\; &= 0.946 M.\end{aligned}\)

We can compute the \({K_b}\) value now that we have all of the equilibrium concentrations:

\(\begin{aligned}{K_b} &= \frac{{\left( {N_4^ + } \right) \times \left( {O{H^ - }} \right)}}{{\left( {N{H_3}} \right)}}\\ &= \frac{{4.093 \times 1{0^{ - 3}} \times 4.093 \times 1{0^{ - 3}}}}{{0.946}}\\ &= 1.771 \times 1{0^{ - 5}}.\end{aligned}\)

The \({K_b}\) value for the ammonia is

\(1.771 \times 1{0^{ - 5}}.\)The final solution is \({K_b} = 1.771 \times 1{0^{ - 5}}.\)