Question

The e.m.f. of the cell $\begin{array}{c}Pt{H}_2\\ 1atm\end{array}\left|\begin{array}{c}HOCN\\ 1.3\times {10}^{-3}M\end{array}\right|\left|\begin{array}{c}A{g}^{+}\\ 0.8M\end{array}\right|Ag\left(s\right)$ is 0.982 V. The K${}_a$ for HOCN is :
$A{g}^{+}+e\to Ag\left(s\right);E^o=0.80V$

• A. $3.33\times {10}^{-4}$
• B. $2.22\times {10}^{-4}$
• C. $4.44\times {10}^{-4}$
• D. $1.6\times {10}^{-3}$

Answer 1

Answer: (D)

$1.6\times {10}^{-3}$

$E_H^0=E_{Ag}^0-E_{cell}^0=0.80-0.982=-0.182V$

$E_H^0=\frac{0.0592}{n}log\frac{\left[H^{+}\right]}{\sqrt{P_{H_2}}}$

$-0.182=\frac{0.0592}{1}log\left[H^{+}\right]$

$\left[H^{+}\right]=0.00085M$

The inital concetrations of HOCN, $H^{+}$ and $OC{N}^{-}$ are 0.0013 M, 0 M and 0 M respectively.

The equilibrium concetrations of HOCN, $H^{+}$ and $OC{N}^{-}$ are $0.0013-0.00085=0.00045M$, 0.00085 M and 0.00085 M respectively.

The dissociation constant Ka is $K_a=\frac{[H6+]\left[OC{N}^{-}\right]}{\left[HOCN\right]}=\frac{0.00085\times 0.00085}{0.00045}=0.0016=1.6\times {10}^{-3}$