Question

The concentration of bivalent lead ions in a sample of polluted water that also contains nitrate ions is determined by adding solid sodium sulphate (M = 142) to exactly 500 mL water. Calculate the molarity of lead ions if 0.355 g of sodium sulphate was needed for complete precipitation of lead ions as sulphate.

• A. $1.25\times {10}^{-3}$ M
• B. $2.5\times {10}^{-3}$ M
• C. $5\times {10}^{-3}$ M
• D. None of these

Answer 1

The correct option is C $5\times {10}^{-3}$ M

$P{b}^{2+}+S{O}_4^{2-}$ (from $N{a}_{2}S{O}_4$) $\to PbS{O}_4$
moles of $N{a}_{2}S{O}_4$ req. = moles of $P{b}^{2+}$ ions
$=\frac{0.355}{142}=0.0025$
$\left[P{b}^{2+}\right]=\frac{0.0025}{0.50}=5\times {10}^{-3}M$