Question

Calculate the first dissociation constant of $H_{3}P{O}_4$ if the emf of the cell: $Hg\left|H{g}_{2}C{l}_2(s)\right|KCl($ salt $)\Vert H_{3}P{O}_4(0.1)M;Na{H}_{2}P{O}_4(0.1)\mid H_2(1atm)pt$ is $-0.3665V.E_{\text{red }}^{\circ }$ of $SHE=0.2412V$

• A. $7.583\times 10^{-5}$
• B. $6.583\times 10^{-5}$
• C. $7.03\times 10^{-5}$
• D. $7.583\times 10^{-3}$

Answer 1

Answer: (D)

$E_{\text{cell }}=-0.3665V$
$E_{SHE}=0.2412V$
$E_{\text{cell }}=-0.2412+x=-0.3665$
$x=-0.3665+0.2412$
$=-0.1253$ For hydrogen half-cell:
$-0.1253=\frac{0.0591}{2}\log {\left[H^{+}\right]}^2$
$\log {\left[H^{+}\right]}^2=-\frac{0.1253\times 2}{0.0591}=-4.2402$
$\left[H^{+}\right]=7.583\times 10^{-3}M$
The hydrogen half-cell is a buffer of $H_{3}P{O}_4$ and $H_{2}P{O}_4^{-}$
$\therefore \left[H^{+}\right]=K_{a_1}\frac{[\text{ acid }]}{[\text{ salt }]}\frac{[\text{ acid }]}{[\text{ salt }]}=1$
$\left[H^{+}\right]=K_{a_1}$
$K_{a_1}=7.583\times 10^{-3}$