At 25^{circ}C, 20.0 mL of 0.2 M weak monopr | vedclass

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(B) At the start, the solution contains only the weak acid $HX$. The concentration $C = 0.2 \ M$. The $pH$ of a weak acid is given by $pH = 0.5(pK_a -

Vedclass · Accepted Answer

$2.0$, $3.3$

Vedclass · Answer

(B) At the start, the solution contains only the weak acid $HX$. The concentration $C = 0.2 \ M$. The $pH$ of a weak acid is given by $pH = 0.5(pK_a - \log C)$. Substituting the values: $pH = 0.5(3.3 - \log 0.2) = 0.5(3.3 - (-0.699)) \approx 0.5(4.0) = 2.0$.$(b)$ When $10 \ mL$ of $0.2 \ M$ $NaOH$ is added to $20 \ mL$ of $0.2 \ M$ $HX$, we reach the half-equivalence point because the moles of $NaOH$ added $(10 \ mL 	imes 0.2 \ M = 2 \ mmol)$ are exactly half the initial moles of $HX$ $(20 \ mL 	imes 0.2 \ M = 4 \ mmol)$. At the half-equivalence point, the concentration of the acid $[HX]$ equals the concentration of its conjugate base $[X^-]$. According to the Henderson-Hasselbalch equation, $pH = pK_a + \log([X^-]/[HX])$. Since $[X^-] = [HX]$, $pH = pK_a = 3.3$.

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