Question

The solubility of $AgCl$ is $0.0015g/L$. The solubility product of $AgCl$ will be:

• A. $2\times 10^{-10}$
• B. $1.1\times 10^{-10}$
• C. $3.1\times 10^{-10}$
• D. $4.1\times 10^{-10}$

Answer 1

Answer: (B)

Find the relationship between $K_{sp}$ and solubility after writing reaction for
dissociation of $AgCl$.
Now solve the problem.
Given: solubility of $AgCl=0.0015g/L$
$=xg/LAgClA{g}^{+}+C{l}^{-}$
After dissociation $xxx$
$K_{sp}=A{g}^{+}]\left[C{l}^{-}\right]$
$=x\times x{K}_{sp}=x^2$
$x=0.0015g/L=\frac{0.0015}{143.5}mol/L$
$\therefore K_{sp}{\left(\frac{0.0015}{143.5}\right)}^2$
$=1.1\times 10^{-10}$