Question

In order to prepare a buffer of $pH=8.26$, the amount of $(N{H}_4{)}_{2}S{O}_4$ required to be mixed with one litre of $0.1\left(M\right)N{H}_3\left(aq\right),$ $p{K}_b$ = $4.74$ is:

• A. $1.0mole$
• B. $10.0mole$
• C. $0.50mole$
• D. $5mole$

Answer 1

Answer: (C)

$0.50mole$

$pOH={pK}_b+log\frac{\left[salt\right]}{\left[base\right]}$

$\Rightarrow 14-pH=4.74+log\frac{\left[salt\right]}{0.1}$

$\Rightarrow 14-8.26=4.74+log\frac{\left[salt\right]}{0.1}$

$\Rightarrow \frac{\left[salt\right]}{0.1}={10}^1=10$

$\Rightarrow \left[salt\right]=1$

Since $1$ mole of ${\left({NH}_4\right)}_2{SO}_4$ gives two moles of ${NH}_4^{+}$. So, $0.5$ mole of ${\left({NH}_4\right)}_2{SO}_4$ in one litre will be needed

${\left({NH}_4\right)}_2{SO}_4\rightleftharpoons 2{NH}_4^{+}+{SO}_4^{2-}$

$\therefore$ correct answer will be C.