The pK_a of a weak acid (HA) is 4.5. The pOH | vedclass

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(D) For a buffer solution,the Henderson-Hasselbalch equation is given by: $pH = pK_a + \log \frac{[A^-]}{[HA]}$Since $50\%$ of the acid is neutralized

Vedclass · Accepted Answer

$9.5$

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(D) For a buffer solution,the Henderson-Hasselbalch equation is given by: $pH = pK_a + \log \frac{[A^-]}{[HA]}$Since $50\%$ of the acid is neutralized,the concentration of the salt $[A^-]$ is equal to the concentration of the remaining acid $[HA]$.Thus,$[A^-] = [HA]$,which means $\frac{[A^-]}{[HA]} = 1$.Substituting the values: $pH = 4.5 + \log(1) = 4.5 + 0 = 4.5$.We know that $pH + pOH = 14$ at $25^{\circ}C$.Therefore,$pOH = 14 - 4.5 = 9.5$.

The $pK_a$ of a weak acid $(HA)$ is $4.5$. The $pOH$ of an aqueous buffered solution of $HA$ in which $50\%$ of the acid is neutralized in its titration with a strong base $(BOH)$ is

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