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)

$BOH+HCl \rightarrow BCl+H_{2}O$
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 $=\frac{2 . 5 \times 2}{5}=1$
$\left[\right.BCl\left]\right.=\frac{1}{10}=0.1M$
For $\left(\right.SAWB\left.\right)$
$pH=7-\frac{1}{2}pk_{b}-\frac{1}{2}logC$
$=7-\frac{1}{2}\times 12-\frac{1}{2}log\left(\left(10\right)^{- 1}\right)$
$=7-6+0.5$
$=1.5$
$\left(H^{+}\right)=3.2\times \left(10\right)^{- 2}$