Question

$2.5mL$ of $\frac{2}{5}M$ weak mono acidic base ( $K_b=1\times 10^{-12}$ at $25^{\circ }C$ ) is titrated with $\frac{2}{15}M$ $HCl$ in water at $25^{\circ }C$ . The concentration of $H^{+}$ at equivalent point is : ( $K_w=1\times 10^{-14}$ at $25^{\circ }C$ )

• A. $3.7\times 10^{-13}M$
• B. $3.2\times 10^{-7}M$
• C. $3.2\times 10^{-2}M$
• D. $27\times 10^{-2}M$

Answer 1

Answer: (C)

$BOH+HCl\to BCl+H_{2}O$
At equivalent point :
$Meq$ . of $BOH=Meq$ . of $HCI$
$2.5\times \frac{2}{5}\times 1=V\times \frac{2}{15}$
$V=7.5mL$
$\therefore$ Total volume $=2.5+7.5=10mL$
$\therefore Meq$ . of $BCl$ formed $=\frac{2.5\times 2}{5}=1$
$[BCl]=\frac{1}{10}=0.1M$
For $(SAWB)$
$pH=7-\frac{1}{2}p{k}_b-\frac{1}{2}logC$
$=7-\frac{1}{2}\times 12-\frac{1}{2}log\left({\left(10\right)}^{-1}\right)$
$=7-6+0.5$
$=1.5$
$\left(H^{+}\right)=3.2\times {\left(10\right)}^{-2}$