Question

Maximum number of moles of $PbS{O}_4$ that can be precipitated by mixing $20.0$ mL of $0.1$ M $Pb(N{O}_3{)}_2$ and $30.0$ mL of $0.1$ M $N{a}_{2}S{O}_4$ will be :

• A. $0.002$
• B. $0.003$
• C. $0.005$
• D. $0.001$

Answer 1

The correct option is A $0.002$

$20.0$ mL of $0.1$ M $P{b}^{2+}$ corresponds to $0.1\times \frac{20.00}{1000}=2\times {10}^{-3}$ moles.
$30.0$ mL of $0.1$ M $S{O}_4^{2-}$ corresponds to $0.1\times \frac{30.00}{1000}=3\times {10}^{-3}$ moles.
The reaction for the formation of lead sulfate is as shown below:
$P{b}^{2+}+{S{O}_4}^{2-}\to PbS{O}_4$
$P{b}^{2+}$ is a limiting quantity, $1$ mole of $P{b}^{2+}$ corresponds to one mole of lead sulphate.
Hence, $0.002$ moles of $P{b}^{2+}$ will form $0.002$ moles of lead sulfate.