Calculate the pH of 100.0 mL of a buffer that is 0.0700 M NH4CI and 0.110 M NH3 before and after the addition of 1.00 mL of 5.90 M HNO3.

Henderson Hasselbalch equation

pH of a buffer is calculated using  Henderson Hasselbalch equation.

For a basic buffer, 

$$\begin{gathered} \mathrm{pOH}={\mathrm{pK}}_{\mathrm{b}}+\log \frac{\left[\mathrm{salt}\right]}{\left[\mathrm{base}\right]} \\ \left[\mathrm{salt}\right]=\mathrm{concentration} \mathrm{of} \mathrm{salt} \\ \left[\mathrm{base}\right]=\mathrm{concentration} \mathrm{of} \mathrm{base} \end{gathered}$$

pH before addition

$\begin{array}{rcl}\mathrm{pOH}& =& {\mathrm{pK}}_{\mathrm{b}}+\log \frac{\left[\mathrm{salt}\right]}{\left[\mathrm{base}\right]}\\ & & \\ {\mathrm{K}}_{\mathrm{b}}& =& 1.78\times {10}^{-5}\\ & & \\ \mathrm{pKb}& =& -\log \left({\mathrm{K}}_{\mathrm{b}}\right)\\ & & \\ & =& -\log (1.78\times {10}^{-5})\\ & & \\ & =& 4.74\\ & & \\ \left[\mathrm{salt}\right]& =& 0.07 \mathrm{M}\\ & & \\ \left[\mathrm{base}\right]& =& 0.11 \mathrm{M}\\ & & \\ & & \mathrm{On} \mathrm{substitution},\\ & & \\ \mathrm{pOH}& =& {\mathrm{pK}}_{\mathrm{b}}+\log \frac{\left[\mathrm{salt}\right]}{\left[\mathrm{base}\right]}\\ & & \\ & =& 4.74+\log \frac{0.07}{0.110}\\ & & \\ & =& 4.54\\ & & \\ \mathrm{pH}+\mathrm{pOH}& =& 14\\ & & \\ \mathrm{pH}& =& 14-\mathrm{pOH}\\ & & \\ & =& 14-4.54\\ & & \\ \mathrm{pH}& =& 9.46\end{array}$

On adding 1mL 5.9 M HNO₃,

$\mathrm{concentration} \mathrm{of} {\mathrm{HNO}}_3=5.90 \mathrm{M}$

On adding the acid, the ammonia reacts with the H⁺ ions.

${\mathrm{NH}}_3+{\mathrm{H}}_3{\mathrm{O}}^{+}\to {\mathrm{NH}}_4\mathrm{OH}+{\mathrm{H}}_2\mathrm{O}$

The concentration in terms of number of moles are;

$\begin{array}{rcl}\mathrm{number} \mathrm{of} \mathrm{moles} \mathrm{of} {\mathrm{NH}}_3& =& \mathrm{molarity}\times \mathrm{total} \mathrm{volume}\\ & & \\ & =& 0.07\times 100\\ & & \\ & =& 7\end{array}$

$$\begin{gathered} {\mathrm{NH}}_3 + {\mathrm{H}}_3{\mathrm{O}}^{+}\to {{\mathrm{NH}}_4}^{+ } +{\mathrm{H}}_2\mathrm{O} \\ \mathrm{I} 7 5.9 10 \\ \mathrm{C} -5.9 -5.9 +5.9 \\ \mathrm{E} 1.18 0 15.9 \end{gathered}$$