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Calculate the $\mathrm{pH}$ of a solution formed by mixing equal volumes of two solutions $\mathrm{A}$ and $\mathrm{B}$ of a strong acid having $\mathrm{pH}=6$ and $\mathrm{pH}=$ respectively.
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Given, $\mathrm{pH}$ of solution $\mathrm{A}=6$
$\left[\mathrm{H}^{+}\right]$of solution $\mathrm{A}=10^{-6} \mathrm{~mol} /$ lit.
$\mathrm{pH}$ of solution $\mathrm{B}=4$
$\left[\mathrm{H}^{+}\right]$of solution $\mathrm{B}=10^{-4} \mathrm{~mol} / \mathrm{lit}$.
On mixing $1 \mathrm{~L}$ of each solution we will get total $2 \mathrm{~L}$ of solution.
Amount of $\left[\mathrm{H}^{+}\right]$in $2 \mathrm{~L}:$ solution $\mathrm{A}=\frac{10^{-6}}{2}$
solution $\mathrm{B}=\frac{10^{-4}}{2}$
Total $\left[\mathrm{H}^{+}\right]$in solution $=\frac{10^{-6}}{2}+\frac{10^{-4}}{2}$
$=10^{-4} \frac{(1+0.01)}{2}=10^{-4} \times 1.01 / 2$
$=5 \times 10^{-5} \mathrm{~mol} / \mathrm{L}$
$\mathrm{pH}=-\log \left[\mathrm{H}^{+}\right]=-\log \left[5 \times 10^{-5}\right]$
$=-\log (5)+(-5 \log 10)$
$=-\log 5+5=4.3$
The $\mathrm{pH}$ of the solution formed by mixing will be $4$ .
$\left[\mathrm{H}^{+}\right]$of solution $\mathrm{A}=10^{-6} \mathrm{~mol} /$ lit.
$\mathrm{pH}$ of solution $\mathrm{B}=4$
$\left[\mathrm{H}^{+}\right]$of solution $\mathrm{B}=10^{-4} \mathrm{~mol} / \mathrm{lit}$.
On mixing $1 \mathrm{~L}$ of each solution we will get total $2 \mathrm{~L}$ of solution.
Amount of $\left[\mathrm{H}^{+}\right]$in $2 \mathrm{~L}:$ solution $\mathrm{A}=\frac{10^{-6}}{2}$
solution $\mathrm{B}=\frac{10^{-4}}{2}$
Total $\left[\mathrm{H}^{+}\right]$in solution $=\frac{10^{-6}}{2}+\frac{10^{-4}}{2}$
$=10^{-4} \frac{(1+0.01)}{2}=10^{-4} \times 1.01 / 2$
$=5 \times 10^{-5} \mathrm{~mol} / \mathrm{L}$
$\mathrm{pH}=-\log \left[\mathrm{H}^{+}\right]=-\log \left[5 \times 10^{-5}\right]$
$=-\log (5)+(-5 \log 10)$
$=-\log 5+5=4.3$
The $\mathrm{pH}$ of the solution formed by mixing will be $4$ .
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