Question

Calculate the amount of $(N{H}_4{)}_{2}S{O}_4$ in g which must be added to $500$ mL of $0.2$ M $N{H}_3$ to yield a solution of $pH=9.35$ $K_b$ for $N{H}_3=4.7$

Answer 1

• Molarity of $N{H}_3$ solution = $\frac{\text{moles of}N{H}_3}{\text{volume of solution}}$

$\text{moles of}N{H}_3=0.2\times 0.5=0.1$

• $pOH=14-pH$
  

$pOH=14-9.35=4.65$

• $p{K}_b=-log\left[K_b\right]=-log[1.78\times {10}^{-5}]=4.74$
• $pOH=p{K}_b-log\frac{\left[base\right]}{\left[salt\right]}$

$log\frac{\left[base\right]}{\left[salt\right]}=4.74-4.65=0.09$

$\frac{\left[base\right]}{\left[salt\right]}=antilog\left[0.09\right]=1.23$

[base] = moles of $N{H}_3=0.1$

$\left[salt\right]=\frac{0.1}{1.23}=0.08$

• moles of ${\left(N{H}_4\right)}^{+}=0.08$
• ${\left(N{H}_4\right)}_{2}S{O}_4\rightleftharpoons {2N{H}_4}^{+}+{\left(S{O}_4\right)}^{-2}$

$\text{moles of}{\left(N{H}_4\right)}_{2}S{O}_4=\frac{\text{moles of}{N{H}_4}^{+}}{2}=0.04$
$\text{amount of}{\left(N{H}_4\right)}_{2}S{O}_4=0.04\times 132=5.28g$