Question

Solubility of $AgCl$ at ${20}^{\circ }$C is $1.435\times {10}^3$ g/L. The solubility product of $AgCl$ is: $[Ag=108,Cl=35.5]$

• A. $1\times {10}^5$
• B. $1\times {10}^{10}$
• C. $1.435\times {10}^5$
• D. $108\times {10}^3$

Answer 1

The correct option is C $1.435\times {10}^5$

For a binary salt,

Solubility= $\sqrt{K_{sp}}$

$\Rightarrow K_{sp}=(solubility{)}^2$

$=(1.435\times {10}^3{)}^2$

$=2.059\times {10}^6$

Solubility product of $AgCl=\frac{2.059\times {10}^6}{143.5}$

$=0.01435\times {10}^6$

$=1.435\times {10}^4$ .