Question

How many moles of ${\left({NH}_4\right)}_2{SO}_4$ must be added to $500mL$ of $0.2M$ ${NH}_3$ to yield a solution of $pH=8.3$ (Given $p{K}_o$ of ${NH}_3=4.7$)

• A. $5\times {10}^{-3}$
• B. $0.5$
• C. $6.6$
• D. $1.32$

Answer 1

The correct option is B $0.5$

Given,

$(N{H}_4{)}_{2}S{O}_4⟶N{H}_{4}HS{O}_4+N{H}_3$

$p{K}_b=4.7,pH=8.3$

Also, Volume of $N{H}_3=500ml$

Molarity of $N{H}_3=0.2M$

Also, $pH=8.3\Rightarrow pOH=14-8.3=5.7$ [$\because pH+pOH=14$]

Molarity of base=$0.2\times \frac{500}{1000}=0.1M$

Now, we know

$pOH=p{K}_b+\log \frac{\left[Salt\right]}{\left[base\right]}$ [$\because$ here, Salt=$(N{H}_4{)}_{2}S{O}_4$, Base=$N{H}_3$]

$\Rightarrow 5.7=4.7+\log \frac{\left[Salt\right]}{\left[0.1\right]}$

$\Rightarrow 1=\log \frac{\left[Salt\right]}{0.1}$

Antilog $1=\frac{\left[Salt\right]}{0.1}\Rightarrow 10=\frac{\left[Salt\right]}{0.1}$

$\Rightarrow \left[Salt\right]=(N{H}_4{)}_{2}S{O}_4=10\times 0.1=1M$

$\Rightarrow$ Number of moles=$Molarity\times Volume$

$\Rightarrow 1\times 0.5L=0.5$ moles.