Question

The $pH$ of the solution obtained by mixing equal volume of solution of $pH=3$ and $pH=4$ is:
[Given: $log55=1.74$]

• A. $2.47$
• B. $3.67$
• C. $3.26$
• D. $4.12$

Answer 1

The correct option is C $3.26$

$pH=3$
$\Rightarrow \left[H^{+}\right]={10}^{-3}M$

$pH=4$
$\Rightarrow \left[H^{+}\right]={10}^{-4}$
Let the initial volume of each sample be $V$.
So the total volume after mixing becomes $2V$.

Total $\left[H^{+}\right]$ on mixing $=\left[\frac{{10}^{-3}+{10}^{-4}}{2}\right]$
$={10}^{-3}\left[\frac{1+{10}^{-1}}{2}\right]$
$={10}^{-3}\left(\frac{1.1}{2}\right)$
$\left[H^{+}\right]=0.55\times {10}^{-3}$

$\therefore pH=-log\left[H^{+}\right]$
$=-log(0.55\times {10}^{-3})$
$=2log10+log55+3log10$
$=2-1.74+3=3.26$