91Ó°ÊÓ

We can utilise the Henderson-Hasselbach equation, which is derived from this reaction, to solve this problem.

\({\rm{N}}{{\rm{H}}_3}(aq) + {{\rm{H}}_2}{\rm{O}}(l) \to {\rm{NH}}_4^ + (aq) + O{{\rm{H}}^ - }(aq)\)

\({K_b}\)can be written as follows:

\({K_b} = \frac{{c\left( {NH_4^ + } \right) \times c\left( {O{H^ - }} \right)}}{{c\left( {N{H_3}} \right)}}\)

Because there is also salt in the solution, \({\rm{N}}{{\rm{H}}_4}{\rm{N}}{{\rm{O}}_3}\), it dissociates:

\({\rm{N}}{{\rm{H}}_4}{\rm{N}}{{\rm{O}}_3}(aq) \to {\rm{NH}}_4^ + (aq) + {\rm{NO}}_3^ - (aq)\)

Now we'll suppose that all of the \({\rm{NH}}_4^ + \) in the \({K_b}\) equation comes from the salt \({\rm{N}}{{\rm{H}}_4}{\rm{N}}{{\rm{O}}_3}\)dissociation. We'll also assume that the \({\rm{N}}{{\rm{H}}_3}\)'s equilibrium concentration is the same as its initial concentration (that is given in the task). The \({K_b}\)equation is as follows:

\({K_b} = \frac{{c\left( {{\rm{N}}{{\rm{H}}_4}{\rm{N}}{{\rm{O}}_3}} \right) \times c\left( {{\rm{O}}{{\rm{H}}^ - }} \right)}}{{c\left( {{\rm{N}}{{\rm{H}}_3}} \right)}}\)

We are given the concentration of \({\rm{O}}{{\rm{H}}^ - }\)and the concentration of \({\rm{N}}{{\rm{H}}_3}\) in the task. Its \({K_b}\)value is \(1.8 \cdot {10^{ - 5}}\), according to the data in Appendix I of the book.

\(\begin{align}c\left( {{\rm{N}}{{\rm{H}}_4}{\rm{N}}{{\rm{O}}_3}} \right) &= \frac{{1.8 \times {{10}^{ - 5}} \times 0.200}}{{1.0 \times {{10}^{ - 5}}}}\\c\left( {{\rm{N}}{{\rm{H}}_4}{\rm{N}}{{\rm{O}}_3}} \right) &= 0.36{\rm{M}}\end{align}\)