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Q.
$pH$ of an aqueous solution of $0.6 \,M \,NH _3$ and $0.4 \,M \,NH _4 Cl$ is $9.4\left( p K_b=4.74\right)$. The new pH when $0.1$ $M \,Ca ( OH )_2$ solution is added to it
Equilibrium
Solution:
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 $]=\left[ NH _3\right]=0.6+0.2=0.8\, M$
$[$ Salt $]=\left[ NH _4 Cl \right]=0.4+0.2=0.2\, M$
$\therefore pOH = p K_b+\log \left[\frac{\text { Salt }}{\text { Base }}\right]$
$ pOH =4.74+\log \left(\frac{0.2}{0.8}\right)$
$=4.74-2 \log 2=4.74-2 \times 0.30=4.14$
$pH =14-4.14=9.86$
