The pH of a 0.01 M weak acid HX (K_{a}=4 tim | vedclass

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Question

(D) For a weak acid $HX$,the dissociation is $HX \rightleftharpoons H^{+} + X^{-}$.Given $C = 0.01 \ M$ and $pH = 5$,so $[H^{+}] = 10^{-5} \ M$.Using

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$25$

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(D) For a weak acid $HX$,the dissociation is $HX ightleftharpoons H^{+} + X^{-}$.Given $C = 0.01 \ M$ and $pH = 5$,so $[H^{+}] = 10^{-5} \ M$.Using $[H^{+}] = \sqrt{K_{a} 	imes C}$,we check: $\sqrt{4 	imes 10^{-10} 	imes 10^{-2}} = \sqrt{4 	imes 10^{-12}} = 2 	imes 10^{-6} \ M$.Since the given $[H^{+}] = 10^{-5} \ M$ does not match the calculated value,the problem statement is internally inconsistent.However,assuming the dilution follows the relation $[H^{+}] = \sqrt{K_{a} 	imes C'}$ for the new concentration $C'$:$10^{-6} = \sqrt{4 	imes 10^{-10} 	imes C'}$$10^{-12} = 4 	imes 10^{-10} 	imes C'$$C' = \frac{10^{-12}}{4 	imes 10^{-10}} = 0.25 	imes 10^{-2} = 25 	imes 10^{-4} \ M$.Thus,$x = 25$.

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