Question

Calculate the $pH$ of a solution formed my mixing equal volumes of two solutions $A$ and $B$ of a strong acid having $pH=6$ and $pH=4$ respectively.

Answer 1

On mixing, the volume gets doubled

$\therefore$ Volume$=2L$

$pH=-log\left[H^{+}\right]$

For solution $A,\left[H^{+}\right]={10}^{-pH}={10}^{-6}mol/d{m}^{-3}$

For solution $B,\left[H^{+}\right]={10}^{-pH}={10}^{-4}mol/d{m}^{-3}$

Now, total $H^{+}moles={10}^{-6}+{10}^{-4}$

$={10}^{-4}({10}^{-2}+1)$

$={10}^{-4}\left(1.01\right)$

After mixing, the volume gets doubled. So, concentration of $H^{+}$ is halved.

$\therefore \left[H^{+}\right]=\frac{1.01\times {10}^{-4}}{2}=0.5\times {10}^{-4}=5\times {10}^{-5}mol/L$

$\therefore pH$ of the resultant solution $=-log\left[H^{+}\right]$

$=-log(5\times {10}^{-5})$

$=5-log5=4.3$

Thus, the $pH$ of resulting solution is $4.3$.