Question

A $1.85g$ sample of an arsenic containing pesticide was chemically converted to As $O_4^{3-}$ (atomic mass of $As=74.9$) and titrated with $P{b}^{2+}$ to form $P{b}_3(As{O}_4{)}_2$. If $20mL$ of $0.1MP{b}^{2+}$ is required to reach the equivalence point, the mass percentage of arsenic in the pesticide sample is closest to

• A. 8.1
• B. 2.3
• C. 5.4
• D. 3.6

Answer 1

Answer: (C)

$3P{b}^{2+}+2As{O}_4\to p{b}_3{\left(As{O}_4\right)}_2$
$3$moles of $P{b}^{2+}$ reacts with $2$ moles $As{O}_4.$
$\therefore 1$ mole of $Pb$ will react with
$=\frac{2}{3}$moles of $As{O}_4,$
Normality $=$ Molarity $\times$ Volume
$N_{p{b}^{2+}}=0.1\times \frac{20}{1000}=2\times 10^{-3}$
$\therefore N_{AS{O}_4}=\frac{2}{3}\times 2\times 10^{-3}=0.00133$
$N_{As}=N_{As{O}_4^{3-}}=0.00133$
$W_{As}=N_{As}\times Mas{s}_{As}$
$=0.00133\times 749$
$=0.0996$
$\%$ of As $=\frac{0.0996}{1.85}\times 100$
$\approx 5.38$
$\approx 5.4\%$