The concentration in M of OH^{-} in 0.001 M | vedclass

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Question

(C) $H_2SO_4$ is a strong diprotic acid, so it dissociates completely as: $H_2SO_4 ightarrow 2H^{+} + SO_4^{2-}$.Concentration of $[H^{+}] = 2 	ime

Vedclass · Accepted Answer

$5 	imes 10^{-12}$

Vedclass · Answer

(C) $H_2SO_4$ is a strong diprotic acid, so it dissociates completely as: $H_2SO_4 ightarrow 2H^{+} + SO_4^{2-}$.Concentration of $[H^{+}] = 2 	imes 0.001 \ M = 0.002 \ M = 2 	imes 10^{-3} \ M$.Using the ionic product of water at $25^{\circ}C$, $K_w = [H^{+}][OH^{-}] = 10^{-14}$.$[OH^{-}] = \frac{K_w}{[H^{+}]} = \frac{10^{-14}}{2 	imes 10^{-3}}$.$[OH^{-}] = 0.5 	imes 10^{-11} \ M = 5 	imes 10^{-12} \ M$.

The concentration in $M$ of $OH^{-}$ in $0.001 \ M$ $H_2SO_4$ is

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