Question

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

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

Answer 1

The correct option is D $0.6\times {10}^{-3}$

Given, $\left[Pb{I}_2\right]=1.5\times {10}^{-3}$ $mol{L}^{-1}$

The reaction involved here is

$Pb{I}_2\left(s\right)\rightleftarrows P{b}^{2+}\left(aq\right)+2I^{-}\left(aq\right)$

At $t=0$ $1.5\times {10}^{-3}$ $0$ $0$

At equilibrium $1.5\times {10}^{-3}-x$ $x$ $2x$

$\therefore$ According to law of conservation of mass

$1.5\times {10}^{-3}-x=x+2x$

$1.5\times {10}^{-3}=4x$

$\therefore x=0.375\times {10}^{-3}$

$\therefore x=$ Concentration of $P{b}^{2+}$ & $2x=$ Concentration of $I^{-}$

$\therefore 2x=0.6\times {10}^{-3}$

$\therefore$ Concentration of Iodide ion=$0.6\times {10}^{-3}g{L}^{-1}$