Question

The solubility of $ AgCl $ is $0.0015 \,g/L$. 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 $K_{s p}$ and solubility after writing reaction for
dissociation of $AgCl$.
Now solve the problem.
Given: solubility of $AgCl =0.0015 \,g / L$
$=x \,g / L\, AgCl \,Ag ^{+}+ Cl ^{-}$
After dissociation $x\, x\, x$
$\left.K_{s p}= Ag ^{+}\right]\left[ Cl ^{-}\right]$
$=x \times x K _{s p}=x^{2}$
$x=0.0015\, g / L =\frac{0.0015}{143.5} \,mol / L$
$\therefore K_{s p}\left(\frac{0.0015}{143.5}\right)^{2}$
$=1.1 \times 10^{-10}$