Question

A weak base $BOH$ of concentration $0.02$ mol ${litre}^{-1}$ has a $pH$ of $10.45$. If $100mL$ of thsi base are mixed with $10mL$ of $0.1M$ $HCl$, what will be the $pH$ (write only integer part) of mixture?

Answer 1

The number of millimoles of HCl is $10mL\times 0.1M=1millimole$
The number of millimoles of base are $0.02M\times 100M=2millimole$
Thus out of 2 millimoles of base, 1 will be neutralized by HCl to form 1 millimole of salt. Thus 1 millimole of base will remain.
Hence, $log\frac{salt}{base}=log\frac{1}{1}=0$.
Hence, the pOH of the buffer solution will be equal to pKb of base.

	Initial concentration (M)	Equilibrium concentration (M)
BOH	0.02	0.02-x
B^+	0	x
OH^-	0	x

$pH=10.45,pOH=14-10.45=3.55$
Hence, $x=\left[O{H}^{-}\right]={10}^{-pOH}={10}^{-3.55}=2.818\times {10}^{-4}$
The equilibrium constant expression is $K=\frac{\left[B^{+}\right]\left[O{H}^{-}\right]}{\left[BOH\right]}$
Substitute values in the above expression.
$K=\frac{(2.818\times {10}^{-4})\times (2.818\times {10}^{-4})}{0.02-(2.818\times {10}^{-4})}=3.97\times {10}^{-6}$
Hence, $p{K}_b=-logK=-log3.97\times {10}^{-6}=5.40$
$pOH=p{K}_b=5.40$
$pH=14-p{K}_b=14-5.40=8.598$

Answer is 8.