Question

Calculate the $pH$ of a solution made by adding $0.01$ mole of $HCl$ in $100mL$ of a solution which is $0.2M$ in $N{H}_3(p{K}_b=4.74)$ and $0.3M$ in $N{H}_4^{+}$:
(Assuming no change in volume)

• A. $5.34$
• B. $8.66$
• C. $7.46$
• D. None of these

Answer 1

The correct option is B $8.66$

$pOH=p{K}_b+\log \frac{\left[\text{Conjugate Acid}\right]}{\left[Base\right]}$
after adding 0.1 mole of HCl, it will react with ammonia and forms conjugate base.
So now, $pOH=p{K}_b+\log \frac{\left[\text{Conjugate Acid}\right]}{\left[Base\right]}=4.74+log(0.3+0.1)/(0.2-0.1)=4.74+log\left(4\right)=5.34$
$pH=14-pOH=8.66$