Question

Calculate the first dissociation constant of $H _{3} PO _{4}$ if the emf of the cell: $Hg \left| Hg _{2} Cl _{2}(s)\right| KCl ($ salt $) \| H _{3} PO _{4}(0.1) M ; NaH _{2} PO _{4}(0.1) \mid H _{2}(1 atm ) pt$ is $-0.3665\, V . E _{\text {red }}^{\circ}$ of $S H E=0.2412 \,V$

• 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)

$E_{\text {cell }}=-0.3665 \,V$
$ E_{S H E}=0.2412 \,V$
$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 $H _{3} PO _{4}$ and $H _{2} PO _{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}$