Question

A buffer solution, $0.080M$ in ${Na}_2{PO}_4$ and $0.020M$ in ${Na}_3{PO}_4$ is prepared. The electrolytic oxidation of $1.0$ mmol of the organic compound $RNHOH$ is carried out in $100mL$ of the buffer. The reaction is as follows:

$RNHOH+H_{2}O⟶R{NO}_2+4H^{+}+4e$

Calculate the approximate $pH$ of the solution after the oxidation is complete.

• A. $6.19$
• B. $7.81$
• C. $10.34$
• D. $12.45$

Answer 1

The correct option is B $7.81$

According to the reaction, $4$ mmol of $H^{+}$ ion will produce $H^{+}=4$ mmol
$\left[H^{+}\right]=\frac{4\times {10}^{-3}}{0.1}=0.04$

${PO}_4^{-3}+H^{+}⟶H{PO}_4^{2-}$ $\frac{1}{k_3}$
$0.02$ $0.04$ $0.08$
$-$ $0.02$ $0.1$

$H{PO}_4^{2-}+H^{+}⟶H_2{PO}_4^{-}$ $\frac{1}{k_2}$
$0.1$ $0.02$
$0.08$ $-$ $0.02$

So, now they form a buffer solution of $H{PO}_4^{2-}$ and $H_2{PO}_4^{-}$.

$pH=p{k}_2+log\frac{0.08}{0.02}$ $(k_2=6.3\times {10}^{-8})$

$pH=7.2+log4=7.8$