Question

Arsenic in 7,150 g of a herbicide is converted in arsenate ions (AsO4{3-}) and the arsenate ions form a precipitate of Ag3AsO4 with 25,00 ml of AgNO3 solution 0,05000 M. The excess of Ag{+} is titrated with 3,85 ml of a KSCN solution 0,0500 M. Calculate the % concentration of arsenic in the sample.

Answer 1

Question #59988, Chemistry, Other

Arsenic in 7,150 g of a herbicide is converted in arsenate ions ($\mathrm{AsO}_4^{(3-)}$) and the arsenate ions form a precipitate of $\mathrm{Ag_3AsO_4}$ with 25,00 ml of $\mathrm{AgNO_3}$ solution 0,0500 M. The excess of $\mathrm{Ag}^{(+)}$ is titrated with 3,85 ml of a KSCN solution 0,0500 M. Calculate the % concentration of arsenic in the sample.

Answer:

The amount of Arsenic can be determined through a difference of $\mathrm{AgNO_3}$ used for its titration and excess of $\mathrm{Ag}^{(+)}$ bounded by KSCN.

$\mathrm{AgNO_3 + KSCN = AgSCN + KNO_3}$

$\mathrm{v(AgNO_3) = v(KSCN)}$

$\mathrm{v(KSCN) = V(KSCN) \cdot C_M(KSCN) = 3.85 / 1000 \cdot 0.0500 = 0.0002 \, mol}$

$$\mathrm{AsO_4^{-3} \, (aq) + 3 \, AgNO_3 \, (aq) \rightarrow Ag_3AsO_4 \, (s) + 3 \, NO_3^{-1} \, (aq)}$$ $$\mathrm{v(AsO_4^{-3}) = 3 \cdot v(AgNO_3)}$$ $$\mathrm{v(AsO_4^{-3}) = (25.00 / 1000 \cdot 0.0500) \cdot 0.0002 = 0.00125 \cdot 0.0002 = 0.00105 \, mol}$$ $$\mathrm{v = m/M \quad m = vM}$$ $$\mathrm{M(AsO_4^{-3}) = 138.9192 \, g/mol}$$ $$\mathrm{m(AsO_4^{-3}) = 0.00105 \cdot 138.9192 = 0.14586 \, g}$$ $$\%(\mathrm{AsO_4^{-3}}) = (0.14586 / 7.150) \cdot 100 = 2.040\%$$

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