Question

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

• A. $3.0\times {10}^{-3}$
• B. $6.0\times {10}^{-3}$
• C. $0.3\times {10}^{-3}$
• D. $0.6\times {10}^{-6}$

Answer 1

The correct option is A $3.0\times {10}^{-3}$

Consider the reaction

$Pb{I}_2\to P{b}^{2+}+2I^{-}$

that means, one mole of $Pb{I}_2$ will give two moles of Iodide ion.

$\left[I^{-}\right]=2(1.5\times {10}^{-3})mol/L$

$\left[I^{-}\right]=3\times {10}^{-3}mol/L$