Question

Calculate the amount of ${(N{H_4})_2}S{O_4}$ in $g$ which must be added to $500mL$ of $0.2M$ $N{H_3}$ to yield a solution of $pH = 9.35{K_b}$ for $N{H_3}1.78x{10^{ - 5}}$ .

Answer 1

To solve this question, first we will find the moles and molarity of the given compound, then we will conclude the base and salt from the given compound. Then we can find the amount of the given compound.

Complete step by step solution:
To calculate the amount of ${(N{H_4})_2}S{O_4}$ , we should find first the mole of ${(N{H_4})_2}S{O_4}$ .
Now, we will find the molarity of $N{H_3}$ solution:
$\because Molarity\,of\,N{H_3}\,solution = \dfrac{{moles\,of\,N{H_3}}}{{volume\,of\,solution}}$
Now,
$Moles\,of\,N{H_3} = 0.2 \times 0.5 = 0.1$
$pOH = 14 - pH$
$\Rightarrow pOH = 14 - 9.35 = 4.65$
$* p{K_b} = - \log [{K_b}] = - \log [1.78 \times {10^{ - 5}}] = 4.74$
$\therefore pOH = p{K_b} - \log \dfrac{{[base]}}{{[salt]}}$
$\log \dfrac{{[base]}}{{[salt]}} = 4.74 - 4.65 = 0.09$
$\dfrac{{[base]}}{{[salt]}} = anti\log [0.09] = 1.23$
$[base] = moles\,of\,N{H_3} = 0.1$
$[salt] = \dfrac{{0.1}}{{1.23}} = 0.08$
Now,
$moles\,of\,{(N{H_4})^ + } = 0.08$
${(N{H_4})_2}S{O_4} \rightleftharpoons 2N{H_4}^ + + {(S{O_4})^{ - 2}}$
So, $moles\,of\,{(N{H_4})_2}S{O_4} = \dfrac{{moles\,of\,N{H_4}^ + }}{2} = 0.04$
$\therefore Amount\,of\,{(N{H_4})_2}S{O_4} = 0.04 \times 132 = 5.28g$

Note:
The reaction of a salt and water to form an acid and a base is called a hydrolysis reaction. Since acids and bases react to form water and salt (neutralization reactions) hydrolysis reactions are the reverse of neutralization reactions.