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A certain amount of $H_2CO_3$ & $HCl$ are dissolved to form $1$ litre solution. At equilibrium it is found that concentration of $H_2CO_3$ & $CO_3^{-\,-}$ are $0.1\,M$ & $0.01\,M$ respectively. Calculate the $pH$ of solution. Given that for $H_2CO_3$ $K_{a_1} =10^{-5}$ & $K_{a_2} =10^{-8}$
$2$
$4$
$1$
$6$
Solution
$\mathrm{H}_{2} \mathrm{CO}_{3}(\mathrm{aq} .) \rightleftharpoons 2 \mathrm{H}_{(\mathrm{aq})}^{+}+\mathrm{CO}_{3}^{-}(\mathrm{aq})$
$\mathrm{K}_{\text {overall }}=\mathrm{K}_{\mathrm{a}_{1}} \times \mathrm{K}_{\mathrm{a}_{2}}=10^{-5} \times 10^{-8}=10^{-13}$
$\because\left[\mathrm{CO}_{3}^{–}\right]=0.01\, \mathrm{M} \&\left[\mathrm{H}_{2} \mathrm{CO}_{3}\right]=0.1 \,\mathrm{M}$
$\therefore \mathrm{K}_{\text {overall }}=\frac{\left[\mathrm{H}^{+}\right]^{2}\left[\mathrm{CO}_{3}^{-}\right]}{\left[\mathrm{H}_{2} \mathrm{CO}_{3}\right]}=10^{-13}$
$\left[\mathrm{H}^{+}\right]=\sqrt{\frac{10^{-13} \times 0.1}{0.01}}=10^{-6}$
$\therefore \mathrm{pH}=6$
