Question

One litre of a buffer solution is prepared by dissolving $0.6$ mole of ${NH}_3$ and $0.4$ mole of ${NH}_{4}Cl$. What is the pH of the solution? for ${NH}_3$, $K_b=1.85\times {10}^{-5}$.
(i) What is the pH of the buffer after addition of $0.1$ mole of $HCl$.
(ii) What is the pH of the buffer after addition of $0.1$ mole of $NaOH$

Answer 1

$pOH=p{K}_b+\log \frac{\left[Salt\right]}{\left[Base\right]}$

$pOH=5-log\left(1.85\right)+\log \frac{0.4}{0.6}$

$pOH=4.57$

$pH=14-4.57=9.43$

(1)After addition of 0.1 mole of $HCl$,0.1 mole of $N{H}_3$ reacts with 0.1 mole of $HCl$ to form 0.1 mole of $N{H}_{4}Cl$.

So,$\left[N{H}_{4}Cl\right]=0.5$

$\left[N{H}_3\right]=0.5$

$pOH=5-log\left(1.85\right)+\log \frac{0.5}{0.5}$

$pOH=4.748$

$pH=14-4.74=9.26$

(2)After addition of 0.1 mole of $NaOH$,0.1 mole of $N{H}_{4}Cl$ reacts with 0.1 mole of $NaOH$ to form 0.1 mole of $N{H}_3$.

So,$\left[N{H}_{4}Cl\right]=0.3$

$\left[N{H}_3\right]=0.7$

$pOH=5-log\left(1.85\right)+\log \frac{0.3}{0.7}$

$pOH=4.38$

$pH=14-4.38=9.62$