Question

The solubility of $PbI_2$ at $25°C$ is $0.7 \,g \,L^{-1}$. The solubility product of $PbI_2$ at this temperature is (molar mass of $PbI_2R = 461.2 \,g \,mol^{-1}$)

• A. $1.40 \times 10^{-9}$
• B. $0.14 \times 10^{-9}$
• C. $140 \times 10^{-9}$
• D. $14.0 \times 10^{-9}$

Answer 1

Answer: (D)

$PbI_2 \rightarrow \underset{\text{S}}{Pb^{++} } + \underset{\text{2S}}{ 2I^-}$
$K_{sp} = s \times \left(2s\right)^{2} = 4s^{3}$
$= 4 \times \left(\frac{0.7}{461.2}\right)^{3} = 14.0 \times 10^{-9}$