A sample of 100 mL of 0.10 M acid HA (K_a = | vedclass

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(D) The initial moles of $HA = 0.10 \ M 	imes 0.100 \ L = 0.01 \ mol$. Let $V$ be the volume of $KOH$ added in $L$. The moles of $KOH$ added $= 0.2 \

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(D) The initial moles of $HA = 0.10 \ M 	imes 0.100 \ L = 0.01 \ mol$. Let $V$ be the volume of $KOH$ added in $L$. The moles of $KOH$ added $= 0.2 	imes V$. At $pH = 5.00$,$[H^+] = 10^{-5} \ M$. Since $K_a = 10^{-5}$,the Henderson-Hasselbalch equation is: $pH = pK_a + \log(\frac{[Salt]}{[Acid]})$. $5.00 = 5.00 + \log(\frac{[A^-]}{[HA]})$,which implies $\log(\frac{[A^-]}{[HA]}) = 0$,so $[A^-] = [HA]$. This occurs at the half-equivalence point. At the half-equivalence point,moles of $KOH$ added $= \frac{1}{2} 	imes 	ext{initial moles of } HA = 0.005 \ mol$. $0.2 	imes V = 0.005 \ mol \implies V = 0.025 \ L = 25 \ mL$. Wait,checking the options: $0.005 / 0.2 = 0.025 \ L = 25 \ mL$. Since $25 \ mL$ is not an option,let's re-evaluate. Actually,at $pH = pK_a$,the ratio of salt to acid is $1:1$. Initial moles $= 0.01$. Half-neutralized means $0.005 \ mol$ of $KOH$ added. $V = 0.005 / 0.2 = 0.025 \ L = 25 \ mL$. Given the options,there might be a typo in the question or options. However,based on the calculation,$25 \ mL$ is the correct result.

$A$ sample of $100 \ mL$ of $0.10 \ M$ acid $HA$ $(K_a = 1 \times 10^{-5})$ is titrated with standard $0.2 \ M \ KOH$. How many $mL$ of $KOH$ will have to be added when the $pH$ in the titration flask will be $5.00$?

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