A solution of 0.1 M weak base (B) is titrate | vedclass

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(C) From the titration curve,the equivalence point is reached at $V = 6 \ mL$ of $HA$ added.At the half-equivalence point,$V = 3 \ mL$,the concentrati

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$2.85$

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(C) From the titration curve,the equivalence point is reached at $V = 6 \ mL$ of $HA$ added.At the half-equivalence point,$V = 3 \ mL$,the concentration of the weak base $[B]$ is equal to the concentration of its conjugate acid $[BH^{+}]$.According to the Henderson-Hasselbalch equation for a weak base:$pOH = pK_{b} + \log\left(\frac{[BH^{+}]}{[B]}\right)$At the half-equivalence point,$[BH^{+}] = [B]$,so $\log(1) = 0$.Therefore,$pOH = pK_{b}$.From the graph,at $V = 3 \ mL$,the $pH$ is approximately $11.15$.Since $pH + pOH = 14$ at $25^{\circ}C$,we have $pOH = 14 - 11.15 = 2.85$.Thus,$pK_{b} = 2.85$.

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