A solution prepared by dissolving equal number of moles of HOCl and NaOCl is a buffer and ${\text{K}}_{\text{a}}=3.2\times {10}^{-8}$. Then its pH is

$\left[\text{HOC}l\right]:\left[\text{NaOC}l\right]$

$\text{acid\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}salt}$

$\begin{array}{l}{\text{P}}^{\text{H}}={\text{P}}^{\text{ka}}+\log \frac{\left[\text{salt}\right]}{\left[\text{acid}\right]}\\ {\text{P}}^{\text{H}}={\text{P}}^{\text{ka}}=-\text{log\hspace{0.17em}ka}\\ \text{=}-\text{log\hspace{0.17em}\hspace{0.17em}}(3.2\times {10}^{-8})\\ =8\log 10-\log 3.2\\ =8-0.5=7.5\end{array}$