0.01 M HA_{(aq)} is 2% ionized; [OH^{-}] of | vedclass

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Question

(C) The dissociation of the weak acid $HA$ is given by: $HA_{(aq)} \rightleftharpoons H^{+}_{(aq)} + A^{-}_{(aq)}$The degree of ionization $\alpha$ is

Vedclass · Accepted Answer

$5 	imes 10^{-11}$

Vedclass · Answer

(C) The dissociation of the weak acid $HA$ is given by: $HA_{(aq)} ightleftharpoons H^{+}_{(aq)} + A^{-}_{(aq)}$The degree of ionization $\alpha$ is $2\% = 0.02$.The concentration of hydrogen ions is $[H^{+}] = C 	imes \alpha = 0.01 	imes 0.02 = 2 	imes 10^{-4} \ M$.Using the ionic product of water,$K_w = [H^{+}][OH^{-}] = 10^{-14}$ at $25^{\circ}C$.Therefore,$[OH^{-}] = \frac{K_w}{[H^{+}]} = \frac{10^{-14}}{2 	imes 10^{-4}} = 0.5 	imes 10^{-10} \ M = 5 	imes 10^{-11} \ M$.

$0.01 \ M \ HA_{(aq)}$ is $2\%$ ionized; $[OH^{-}]$ of solution is

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