Question

A buffer solution $0.04M$ in $N{a}_{2}HP{O}_4$ and $0.02M$ in $N{a}_{3}P{O}_4$ is prepared. The electrolytic oxidation of $1.0$ milli-mole of the organic compound $RNHOH$ is carried out in $100\text{mL}$ of the buffer. The reaction is
$RNHOH+H_{2}O\to RN{O}_2+4H^{+}+4e^{-}$
The approximate $pH$ of solution after the oxidation is complete is:
[Given: For $H_{3}P{O}_4,p{K}_{a_1}=2.2;p{K}_{a_2}=7.20;p{K}_{a_3}=12,log2=0.30andlog4=0.60$]

• A. $6.90$
• B. $7.20$
• C. $7.5$
• D. None of these

Answer 1

The correct option is C $7.5$

Moles of $H^{+}$ produced $=4\times {10}^{-3}$
$\left[H^{+}\right]=\frac{4\times {10}^{-3}}{0.1}=0.04M$
$\begin{array}{cccccc}& P{O}_4^{3-}(aq.)& +& H^{+}(aq.)& \to & HP{O}_4^{2-}(aq.)\\ \text{Initial milli-moles}& 0.02& & 0.04& & \\ \text{Final milli-moles}& -& & 0.02& & 0.02\end{array}$
$\begin{array}{cccccc}& HP{O}_4^{2-}(aq.)& +& H^{+}(aq.)& \to & H_{2}P{O}_4^{-}(aq.)\\ \text{Initial milli-moles}& 0.06& & 0.02& & 0.02\\ \text{Final milli-moles}& 0.04& & 0& & 0.02\end{array}$
Finally buffer solution of $H_{2}P{O}_4^{-}$ (acid) and $HP{O}_4^{2-}$ (conjugate base) is formed
$pH=p{K}_{a_2}+log\frac{\left[HP{O}_4^{2-}\right]}{\left[H_{2}P{O}_4^{-}\right]}=7.2+log\frac{0.04}{0.02}=7.5$