A solution of formic acid is half-neutralized | vedclass

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(A) The reaction is: $HCOOH + KOH \rightarrow HCOOK + H_2O$. Since the acid is half-neutralized,the concentration of the salt $[HCOOK]$ is equal to th

Vedclass · Accepted Answer

$3.6990$

Vedclass · Answer

(A) The reaction is: $HCOOH + KOH ightarrow HCOOK + H_2O$. Since the acid is half-neutralized,the concentration of the salt $[HCOOK]$ is equal to the concentration of the remaining acid $[HCOOH]$. Using the Henderson-Hasselbalch equation for an acidic buffer: $pH = pK_a + \log \frac{[Salt]}{[Acid]}$. Given $K_a = 2 	imes 10^{-4}$,so $pK_a = -\log(2 	imes 10^{-4}) = 4 - \log 2 = 4 - 0.3010 = 3.6990$. Since $[Salt] = [Acid]$,the ratio $\frac{[Salt]}{[Acid]} = 1$. Therefore,$pH = 3.6990 + \log(1) = 3.6990 + 0 = 3.6990$.

$A$ solution of formic acid is half-neutralized by the addition of a $KOH$ solution. If $K_a (HCOOH) = 2 \times 10^{-4}$,what is the $pH$ of the solution? (Given: $\log 2 = 0.3010$)

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