91Ó°ÊÓ

To calculate the \(pH\)of a buffer solution, use the H H (Henderson-Hasselbach) equation:

\(pH = p{K_a} + \log \left( {\frac{{c({\rm{ base }})}}{{c({\rm{ acid }}))}}} \right)\)

The H H equation for a base and its\(Kb\)value is easier to deduce in this scenario since the buffer is made up of a weak base and its salt.

\(\begin{align}{K_b} &= \frac{{c\left( {O{H^ - }} \right) \cdot c\left( {NH_4^ + } \right)}}{{c\left( {N{H_3}} \right)}}\\{K_b} \cdot c\left( {N{H_3}} \right) &= c\left( {O{H^ - }} \right) \cdot c\left( {NH_4^ + } \right)\\c\left( {O{H^ - }} \right) &= \frac{{{K_b} \cdot c\left( {N{H_3}} \right)}}{{c\left( {NH_4^ + } \right)}}\\pOH &= p{K_b} - \log \left( {\frac{{c\left( {N{H_3}} \right)}}{{c\left( {NH_4^ + } \right)}}} \right)\end{align}\)

We must first compute the concentrations of\({{\mathop{\rm NH}\nolimits} _3}\)and\({{\mathop{\rm NH}\nolimits} _4}^ + \)before plugging the quantities into this equation. We can accomplish so by utilising the following formula:

\(c = \frac{n}{V}\)

where \(n\) denotes the number of moles in the solution and \(v\) denotes the volume of the solution.

\(\begin{align}\left( {N{H_3}} \right) &= \frac{{0.20\;{\rm{mol}}}}{{1.00\;{\rm{L}}}}c\left( {N{H_3}} \right)\\ &= 0.20M\end{align}\)

\(\begin{align}c\left( {NH_4^ + } \right) &= \frac{{0.40\;{\rm{mol}}}}{{1.00\;{\rm{L}}}}\\c\left( {NH_4^ + } \right) &= 0.40{\rm{M}}\end{align}\)

We can now compute the buffer solution's \(pOH\).:

\(\begin{align}pOH &= p{K_b} - \log \left( {\frac{{c\left( {N{H_3}} \right)}}{{c\left( {NH_4^ + } \right)}}} \right)\\pOH &= - \log \left( {1.8 \cdot {{10}^{ - 5}}} \right) - \log \left( {\frac{{0.20}}{{0.40}}} \right)\\pOH &= 5.04\end{align}\)

The \(Kb\) value can be found in the book's Appendix I table. The formula for calculating \(pH\) is as follows:

\(\begin{align}pH &= 14 - pOH\\pH &= 14 - 5.04\\pH &= 8.96\end{align}\)

With these concentrations of \({\rm{N}}{{\rm{H}}_3}\)and \({\rm{N}}{{\rm{H}}_4}{\rm{N}}{{\rm{O}}_3}\), the \(pH\) of the solution is \(8.96\).