Question

A $1.85\, g$ sample of an arsenic containing pesticide was chemically converted to As $O_{4}^{3-}$ (atomic mass of $As = 74.9$) and titrated with $Pb^{2+}$ to form $Pb_{3}(AsO_{4})_{2}$. If $20\, mL$ of $0.1\, M\, Pb^{2+}$ 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)

$3Pb^{2+}+2AsO_{4} \rightarrow pb_{3} \left(AsO_{4}\right)_{2}$
$3$moles of $Pb^{2+}$ reacts with $2$ moles $AsO_{4}.$
$\therefore 1$ mole of $Pb$ will react with
$=\frac{2}{3} $moles of $AsO_{4}, $
Normality $=$ Molarity $\times$ Volume
$N_{pb^{2+}} =0.1\times\frac{20}{1000}=2\times10^{-3}$
$\therefore N_{ ASO _{4}}=\frac{2}{3} \times 2 \times 10^{-3}=0.00133$
$N_{ As }=N_{ AsO _{4}^{3-}}=0.00133$
$W_{ As }=N_{ As } \times Mass _{ As }$
$=0.00133 \times 749$
$=0.0996 $
$\%$ of As $=\frac{0.0996}{1.85} \times100$
$\approx 5.38 $
$\approx 5.4\%$