Question

The EMF of the cell:
$Ag|AgCl,0.1MKCl||0.1MAgN{O}_3|Agis0.45V.0.1M,KCl$ is $85\%$ dissociated and $0.1$ M $AgN{O}_3$ is $82\%$ dissociated. Calculate the solubility product of $AgCl$ at ${25}^0$C.

• A. $3.4\times {10}^{-10}$
• B. $1.7\times {10}^{-10}$
• C. $5.25\times {10}^{-24}$
• D. none of these

Answer 1

Answer: (B)

$1.7\times {10}^{-10}$

$E_{cell}=E_{cell}^o-\frac{591}{n}\log \frac{\left[A{g}^{+}\right]anode}{\left[A{g}^{+}\right]cathode}$

$\left[A{g}^{+}\right]anode=2.055\times {10}^{-9}M$

$K_{sp}=\left[A{g}^{+}\right]\left[C{l}^{-}\right]=(2.055\times {10}^{-9})\left(0.085\right)=0.1746\times {10}^{-9}$

$=1.75\times {10}^{-10}$