What will be the amount of $\left(\mathrm{NH}_4\right)_2 \mathrm{SO}_4$ (in g ) which must be added

Question

What will be the amount of $\left(\mathrm{NH}_4\right)_2 \mathrm{SO}_4$ (in g ) which must be added to 500 mL of $0.2 \mathrm{M} \mathrm{NH}_4 \mathrm{OH}$ to yield a solution of pH 9.35? [Given : $\mathrm{p} K_a$ of $\mathrm{NH}_4^{+}=9.26$, $\mathrm{p} K_b$ of $\left.\mathrm{NH}_4 \mathrm{OH}=14-\mathrm{p} K_a\left(\mathrm{NH}_4^{+}\right)\right]$

Professor PC Thomas Classes · Answer

(a) : $\mathrm{p} K_a$ of $\mathrm{NH}_4^{+}=9.26$
Hence, $\mathrm{pK} K_b$ of $\mathrm{NH}_4 \mathrm{OH}=14-9.26=4.74$
$
\mathrm{pOH}=14-9.35=4.65
$
Each 1 mol of $\left(\mathrm{NH}_4ight)_2 \mathrm{SO}_4$ furnishes 2 moles of $\mathrm{NH}_4^{+}$in solution.
Let, $\left[\left(\mathrm{NH}_4ight)_2 \mathrm{SO}_4ight]=x \mathrm{~mol} \mathrm{~L}^{-1}$
$
\left[\mathrm{NH}_4^{+}ight]=2 	imes \mathrm{mol} \mathrm{~L}^{-1} ;\left[\mathrm{NH}_4 \mathrm{OH}ight]=0.2 \mathrm{~mol} \mathrm{~L}^{-1}
$
Using Henderson-Hasselbalch equation,
$
\mathrm{pOH}=\mathrm{p} K_b+\log \frac{\left[\mathrm{NH}_4^{+}ight]}{\left[\mathrm{NH}_4 \mathrm{OH}ight]} ; 4.65=4.74+\log \left(\frac{2 x}{0.2}ight)
$
This gives $x=0.081 \mathrm{~mol} \mathrm{~L}^{-1}$
$
=0.0405 \mathrm{~mol} 	ext { in } 500 \mathrm{~mL} 	ext { (as required) }=0.0405 	imes 132=5.35 \mathrm{~g}
$