pH of 0.1 M solution of a weak acid (HA) is | vedclass

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(B) For a weak acid $HA$,the dissociation is $HA \rightleftharpoons H^+ + A^-$.Given $[H^+] = 10^{-pH} = 10^{-4.50}$.Using the formula $[H^+] = \sqrt{

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$8$

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(B) For a weak acid $HA$,the dissociation is $HA \rightleftharpoons H^+ + A^-$.Given $[H^+] = 10^{-pH} = 10^{-4.50}$.Using the formula $[H^+] = \sqrt{K_a \cdot C}$,where $C = 0.1 \ M$:$(10^{-4.50})^2 = K_a \cdot 0.1$$10^{-9} = K_a \cdot 10^{-1}$$K_a = 10^{-8}$.$pK_a = -\log(K_a) = -\log(10^{-8}) = 8$.When the acid is neutralized with $NaOH$ to decrease the acid content to half,we form a buffer solution where $[Salt] = [Acid]$.According to the Henderson-Hasselbalch equation: $pH = pK_a + \log \frac{[Salt]}{[Acid]}$.Since $[Salt] = [Acid]$,$pH = pK_a + \log(1) = pK_a$.Therefore,$pH = 8$.

$pH$ of $0.1 \ M$ solution of a weak acid $(HA)$ is $4.50$. It is neutralized with $NaOH$ solution to decrease the acid content to half. The $pH$ of the resulting solution is:

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