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 $P{b}^{2+}$ to form $P{b}_3(As{O}_4{)}_2$. If 20 mL of 0.1 M $P{B}^{2+}$ is required to reach the equivalence point, the mass percentages of arsenic in the pesticide sample is closest to

• A. 8.1
• B. 2.3
• C. 5.4
• D. 3.6

Answer 1

The correct option is B 2.3

Given, Mass of sample=$1.85g$

The reaction involved is

$3P{b}^{2+}+2As{O}_4^{3-}⟶P{b}_3(As{O}_4{)}_2$

Now, Volume of $P{b}^{2+}$ consumed for neutralization=$20ml$

Molarity of $P{b}^{2+}=0.1M$

$\therefore$ No. of moles of $P{b}^{2+}=0.1\times \frac{20}{1000}$

$=2\times {10}^{-3}$ moles

Now, from reaction it is clear that

$3$ moles $P{b}^{2+}$ produce $1$ mole $P{b}_3(As{O}_4{)}_2$

We know $1$ mole $P{b}_3(As{O}_4{)}_2$ has $1$ mole $As$ in it.

$\Rightarrow 3$ moles $P{b}^{2+}$ produce $1$ mole $As$

$\therefore 2\times {10}^{-3}$ moles $P{b}^{2+}$ will give $\frac{2\times {10}^{-3}}{3}$ moles of $As$

$\therefore$ No. of moles of $As$ in the compound=$0.66\times {10}^{-3}$ moles

Now,

We know, $1$ mole $As=74g$

$\Rightarrow 0.66\times {10}^{-3}$ moles $As=0.66\times {10}^{-3}\times 74g$

$=48.84\times {10}^{-3}g$

$=0.0488g$

Now % calculation,

Mass% of $As=\frac{MassofAs}{MassofSample}\times 100$

$=\frac{0.0488}{1.85}\times 100$

$=0.0263\times 100$

$=2.6$% $\approx 2.3$%