A certain amount of H_2CO_3 & HCl are&nbs | vedclass

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Question

$\mathrm{H}_{2} \mathrm{CO}_{3}(\mathrm{aq} .) \rightleftharpoons 2 \mathrm{H}_{(\mathrm{aq})}^{+}+\mathrm{CO}_{3}^{-}(\mathrm{aq})$

$\mathrm{K}_{\

Vedclass · Accepted Answer

$6$

Vedclass · Answer

$\mathrm{H}_{2} \mathrm{CO}_{3}(\mathrm{aq} .) ightleftharpoons 2 \mathrm{H}_{(\mathrm{aq})}^{+}+\mathrm{CO}_{3}^{-}(\mathrm{aq})$

$\mathrm{K}_{	ext {overall }}=\mathrm{K}_{\mathrm{a}_{1}} 	imes \mathrm{K}_{\mathrm{a}_{2}}=10^{-5} 	imes 10^{-8}=10^{-13}$

$\because\left[\mathrm{CO}_{3}^{--}ight]=0.01\, \mathrm{M} \&\left[\mathrm{H}_{2} \mathrm{CO}_{3}ight]=0.1 \,\mathrm{M}$

$	herefore \mathrm{K}_{	ext {overall }}=\frac{\left[\mathrm{H}^{+}ight]^{2}\left[\mathrm{CO}_{3}^{-}ight]}{\left[\mathrm{H}_{2} \mathrm{CO}_{3}ight]}=10^{-13}$

$\left[\mathrm{H}^{+}ight]=\sqrt{\frac{10^{-13} 	imes 0.1}{0.01}}=10^{-6}$

$	herefore \mathrm{pH}=6$

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 $H_2CO_3$ $K_{a_1} =10^{-5}$ & $K_{a_2} =10^{-8}$

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