Two solutions, $A$ and $B$, each of $100 \mathrm{~L}$ was made by dissolving $4 \mathrm{~g}$ of $\mathrm{NaOH}$ and $9.8 \mathrm{~g}$ of $\mathrm{H}_{2} \mathrm{SO}_{4}$ in water, respectively. The $\mathrm{pH}$ of the resultant solution obtained from mixing $40 \mathrm{~L}$ of solution $A$ and $10 \mathrm{~L}$ of solution. $B$ is _________________
(10.60) $\mathrm{M}_{\mathrm{H}_{2} \mathrm{SO}_{4}}=\frac{9.8}{98 \times 100}=10^{-3} \mathrm{M}$
$\mathrm{M}_{\mathrm{NaOH}}=\frac{4}{40 \times 100}=10^{-3} \mathrm{M}$
After neutralisation $\left[\mathrm{OH}^{-}\right]$can be calculated as
$\left[\mathrm{OH}^{-}\right]=\frac{\left(40 \times 10^{-3}\right)-\left(2 \times 10^{-3} \times 10\right)}{50}$
$=\frac{20}{50} \times 10^{-3}$
$\left[\mathrm{OH}^{-}\right]=\frac{2}{5} \times 10^{-3}$
$\mathrm{pOH}=3.397$
$\mathrm{pH}=14-\mathrm{pOH}$
$14-3.397=10.603$
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