Question

A one litre solution contains $0.04M$ $N{a}_{2}HP{O}_4$ and $0.02M$ $Na{H}_{2}P{O}_4$. The correct statement(s) is/are:

For $H_{3}P{O}_4:$ $p{K}_{a_1}=2.14;p{K}_{a_2}=6.84;p{K}_{a_3}=12.3$ and use $log2=0.3$.

• A. pH of solution is $7.14$
• B. On adding $1$ mmol $HCl$ to initial solution pH decreases by more than $0.6$ units
• C. On adding $0.01$ mol of $H_{2}S{O}_4$ to initial solution pH decreases by $0.6$ units
• D. On adding $0.04$ mol $HCl$ to initial solution pH of solution becomes $6.84$

Answer 1

Answer: (A), (C)

pH of solution is $7.14$
On adding $0.01$ mol of $H_{2}S{O}_4$ to initial solution pH decreases by $0.6$ units

$pH=6.84+log\frac{0.04}{0.02}=7.14$

Hence, pH of the solution is $7.14$.

$\underset{\underset{t=\infty }{t=0}}{-}\underset{\underset{0.039}{0.04}}{HP{O}_4^{2-}}+\underset{\underset{0}{{10}^{-3}}}{HCl}\to \underset{\underset{0.021}{0.02}}{H_{2}P{O}_4^{-}}$
$p{H}_f=6.84+log\frac{0.039}{0.021}=p{H}_1$

$\underset{\underset{t=\infty }{t=0}}{-}\underset{\underset{0.02}{0.04}}{2HP{O}_4^{2-}}+\underset{\underset{0}{0.01}}{H_{2}S{O}_4}\to \underset{\underset{0.04}{0.02}}{2H_{2}P{O}_4^{-}}$
$p{H}_f=6.84+log\frac{1}{2}=6.54$
$\Delta pH=-0.6$

Hence, on adding $0.01$ moles of $H_{2}S{O}_4$ tp initial solution, pH decreases by $0.6$ units.

$\underset{\underset{t=\infty }{t=0}}{-}\underset{\underset{0}{0.04}}{HP{O}_4^{2-}}+\underset{\underset{0}{0.04}}{H^{+}}\to \underset{\underset{0.06}{0.02}}{H_{2}P{O}_4^{-}}$
$pH=\frac{p{K}_{a_1}+p{K}_{a_2}}{2}=4.49$

Hece, on adding $0.04$ moles $HCl$ to initial solution, pH of solution becomes $4.49$ and not $6.84$. Hence, option D is incorrect.

Hence, options A and C are correct.