Question

If the molar concentration of $Pb{I}_2$ is $1.5\times {10}^{-3}mol$ $L^{-1}$, the concenration of iodide ions in $g$ ion $L^{-1}$ is:

• A. $3.10\times {10}^{-3}$
• B. $6.0\times {10}^{-3}$
• C. $0.31\times {10}^{-3}$
• D. $0.6\times {10}^{-6}$

Answer 1

The correct option is A $3.10\times {10}^{-3}$

$Pb{I}_{2\left(soln\right)}\rightleftharpoons P{b}^{2+}+2I^{-}{K}_{sp}=\left[P{b}^{2+}\right]{\left[I^{-}\right]}^{2}1.5\times {10}^{-8}=x\times {\left(2x\right)}^2{x}^3=\frac{1.5}{4}\times {10}^{-8}=3.75\times {10}^{-9}x=1.55\times {10}^{-3}$

So, $\left[I^{-}\right]=2(1.55\times {10}^{-3})=3.10\times {10}^{-3}$