Sulphurous acid (H_{2}SO_{3}) has Ka_{1} = 1. | vedclass

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(B) $H_{2}SO_{3}$ is a dibasic acid with concentration $c = 0.588 \ M$. Since $Ka_{1} \gg Ka_{2}$,the $pH$ is primarily determined by the first dissoc

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$1$

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(B) $H_{2}SO_{3}$ is a dibasic acid with concentration $c = 0.588 \ M$. Since $Ka_{1} \gg Ka_{2}$,the $pH$ is primarily determined by the first dissociation step: $H_{2}SO_{3(aq)} ightleftharpoons H^{+}_{(aq)} + HS{O_{3}}^{-}_{(aq)}$ $Ka_{1} = \frac{[H^{+}][HSO_{3}^{-}]}{[H_{2}SO_{3}]} = 1.7 	imes 10^{-2}$ Let $x$ be the concentration of $H^{+}$ at equilibrium: $\frac{x^{2}}{0.588 - x} = 0.017$ $x^{2} + 0.017x - 0.009996 = 0$ Using the quadratic formula $x = \frac{-b + \sqrt{b^{2} - 4ac}}{2a}$: $x = \frac{-0.017 + \sqrt{(0.017)^{2} - 4(1)(-0.009996)}}{2} \approx 0.09186 \ M$ $pH = -\log[H^{+}] = -\log(0.09186) \approx 1.036$ Rounding to the nearest integer,the $pH$ is $1$.

Sulphurous acid $(H_{2}SO_{3})$ has $Ka_{1} = 1.7 \times 10^{-2}$ and $Ka_{2} = 6.4 \times 10^{-8}$. The $pH$ of $0.588 \ M \ H_{2}SO_{3}$ is ..... . (Round off to the Nearest Integer)

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