Question

Two weak acid solutions $H{A}_1$and $H{A}_2$ each with the same concentration and having pKa values 3 and 5 are placed in contact with hydrogen electrode ( 1 atm, 25${}^0$C ) and are interconnected through a salt bridge. Find emf of the cell.

• A. + 0.059 V
• B. 0.059 V
• C. 0.0295 V
• D. 0.118 V

Answer 1

The correct option is A + 0.059 V

The expression for dissociation constant is $HA\stackrel{}{\to }{H}^{+}+A^{-}{K}_a=\frac{\left[H^{+}\right]\left[A^{-}\right]}{\left[HA\right]}=\frac{{\left[H^{+}\right]}^2}{\left[HA\right]}$
The expression for the difference in pKa values for acids will be $HA\stackrel{}{\to }{H}^{+}+A^{-}{pK}_{a2}-{pK}_{a1}=\left[-log{\left\{\frac{{\left[H^{+}\right]}^2}{\left[HA\right]}\right\}}_2\right]-\left[-log{\left\{\frac{{\left[H^{+}\right]}^2}{\left[HA\right]}\right\}}_1\right]=log\frac{{\left[H^{+}\right]}_1^2}{{\left[H^{+}\right]}_2^2}\frac{1}{2}[{pK}_{a2}-{pK}_{a1}]=log\frac{{\left[H^{+}\right]}_1}{{\left[H^{+}\right]}_2}$
The emf of cell is
$E_{cell}=E_{cell}^0+\frac{0.0592}{n}log\frac{{\left[H^{+}\right]}_1}{{\left[H^{+}\right]}_2}=0.0+\frac{0.0592}{1}\frac{1}{2}[{pK}_{a2}-{pK}_{a1}]=0.0+0.0592\frac{[5-3]}{2}=\mathrm{0.0592.}$