Question

$pH$ of an aqueous solution of $0.6MN{H}_3$ and $0.4MN{H}_{4}Cl$ is $9.4\left(p{K}_b=4.74\right)$. The new pH when $0.1$ $MCa(OH{)}_2$ solution is added to it

• A. $9.86$
• B. $10.14$
• C. $10.2$
• D. $10.86$

Answer 1

Answer: (A)

Since $Ca(OH{)}_2$ is completely ionized

BBB Rule: On adding base, to the basic buffer, concentration of base increases and salt decreases
$\therefore$ New concentration of base and salt are:
$[$ Base $]=\left[N{H}_3\right]=0.6+0.2=0.8M$
$[$ Salt $]=\left[N{H}_{4}Cl\right]=0.4+0.2=0.2M$
$\therefore pOH=p{K}_b+\log \left[\frac{\text{ Salt }}{\text{ Base }}\right]$
$pOH=4.74+\log \left(\frac{0.2}{0.8}\right)$
$=4.74-2\log 2=4.74-2\times 0.30=4.14$
$pH=14-4.14=9.86$