Question

5 ml of $0.1MPb(N{O}_3{)}_2$ is mixed with 10 ml of 0.02 M $KI$. The amount of $Pb{I}_2$ precipitated will be about:

• A. ${10}^{-2}mol$
• B. ${10}^{-4}mol$
• C. $2\times {10}^{-4}mol$
• D. ${10}^{-3}mol$

Answer 1

The correct option is B ${10}^{-4}mol$

$\text{5 ml of}0.1MPb(N{O}_3{)}_2=5\times 0.1\times {10}^{-3}=5\times {10}^{-4}mol$
$\text{10 ml of}0.02MKI=10\times 0.02\times {10}^{-3}=2\times {10}^{-4}mol$

$\underset{5\times {10}^{-}4mol}{Pb(N{O}_3{)}_2}\to \underset{5\times {10}^{-}4mol}{P{b}^{2+}}+2N{O}_3^{-}$

$\underset{2\times {10}^{-4}}{KI}\to K^{+}\underset{2\times {10}^{-4}}{I^{-}}$

Again,
$P{b}^{2+}+2I^{-}\to Pb{I}_2$
$\therefore$ Limiting reagent is $I^{-}$, hence amount of $Pb{I}_2$ precipitated will be ${10}^{-4}$ mol.