Question

Calculate the amount of $(N{H}_4{)}_{2}S{O}_4$ (in gram) which must be added to $1L$ solution of $0.2M$ $N{H}_{4}OH$ to yield a solution of $pH=9.35$ at ${25}^{o}C$.
(Assume volume is not changing during the reaction)
$K_b$ for $N{H}_{4}OH$ is $1.8\times {10}^{-5}$
Take $\log 1.8=0.25and{10}^{0.1}=1.25$

• A. $5.33g$
• B. $5.56g$
• C. $8.56g$
• D. $10.56g$

Answer 1

The correct option is D $10.56g$

Molarity of $N{H}_{4}OH$ soluiton $=\frac{\text{moles of}N{H}_{4}OH}{\text{volume of solution}}$
$\text{Moles of}N{H}_{4}OH=\text{Molarity}\times \text{Volume}=0.2\times 1=0.2$

$pOH=14-pH$ at ${25}^{o}C$
$pOH=14-9.35-=4.65$
$p{K}_b=-log\left[K_b\right]=-\log [1.8\times {10}^{-5}]p{K}_b=5-\log 1.8p{K}_b=5-0.25=4.75$
$pOH=p{K}_b-\log \left(\frac{\left[\text{base}\right]}{\left[\text{Cation of salt}\right]}\right)$
$\log \left(\frac{\left[\text{base}\right]}{\left[\text{Cation of salt}\right]}\right)=4.75-4.65=0.1$
$\left(\frac{\left[\text{base}\right]}{\left[\text{Cation of salt}\right]}\right)={10}^{\left[0.1\right]}=1.25$
[base] = moles of $N{H}_{4}OH=0.2$
$\left[\text{Cation of salt}\right]=\frac{0.2}{1.25}=0.16$
Moles of $\left(N{H}_4^{+}\right)=0.16$
$(N{H}_4{)}_{2}S{O}_4\rightleftharpoons 2N{H}_4^{+}+(S{O}_4^{2-})$
Moles of $(N{H}_4{)}_{2}S{O}_4=\frac{molesofN{H}_4^{+}}{2}=0.08$
Amount of $(N{H}_4{)}_{2}S{O}_4=0.08\times 132=10.56g$