Question

2.5 mL of $\frac{2}{5}M$ weak monoacidic base$(K_b=1\times {10}^{-12}at{25}^{\circ }C)$ is titrated with$\frac{2}{15}MHCl$ in water at $25{}^{\circ }C$. The $pH$ at equivalence point is
$(K_w=1\times {10}^{-14}at25{}^{\circ }C)$

• A. 12.5
• B. 1.5
• C. 2
• D. 7.5

Answer 1

The correct option is B 1.5

mmol of base $=2.5\times \frac{2}{5}=1$
mmol of acid required to reach the end point = $\text{1}$
Volume of acid required to reach the end point $=\frac{15}{2}mL$
Total volume at the end point $=\frac{15}{2}+2.5=10mL$
Molarity of salt at the end point$=\frac{1}{10}=0.10$

$pOH=\frac{1}{2}pKw+\frac{1}{2}pKb+\frac{1}{2}logC$
$pOH=7+6+\frac{1}{2}log0.1$
= 12.5
$pH=14-12.5=1.5$