Question

A 0.400 M solution of ammonia was titrated with hydrochloric acid to the equivalence point, where the total volume was 1.50 times the original volume. At what pH does the equivalence point occur?

Short answer

The pH when the equivalence point occurs at4.92.

Step-by-step solution

The concentration of NH4+. 

The molar amount of ammonia can be denoted as,

$\begin{array}{c}{M}_1\times V_1=M_2\times V_2\\ (0.4MNH{}_3)V_{N{H}_3}=0.4VmolN{H}_3\end{array}$

The concentration therefore is,

$\frac{0.4VmolN{H}_4^{+}}{1.5VL}=0.267MN{H}_4^{+}$

The dissociation reaction that goes,

$N{H}_4^{+}\rightleftharpoons N{H}_3+H_{2}O$

The pH at which the equivalent point occurs.

The expression for dissociation constant can be,

$\begin{array}{l}{K}_a=\frac{\left[N{H}_3\right]\left[H_3{O}^{+}\right]}{\left[N{H}_4^{+}\right]}\\ K_a=\frac{K_w}{K_b}=\frac{1.0\times {10}^{-14}}{1.8\times {10}^{-5}}\end{array}$

Therefore,

$\frac{1.0\times {10}^{-14}}{1.8\times {10}^{-5}}=\frac{\left[N{H}_3\right]\left[H_3{O}^{+}\right]}{\left[N{H}_4^{+}\right]}$

Hence,

$\frac{1.0\times {10}^{-14}}{1.8\times {10}^{-5}}=\frac{x.x}{\mathrm{0.267.}x}$

After calculation,

$x=1.2\times {10}^{-5}$

The calculated pH is therefore,

$\begin{array}{l}pH=-\log \left[H^{+}\right]\\ pH=-\log (1.2\times {10}^{-5})\\ pH=4.92\end{array}$