Answer to Question #117377 in Chemistry for Em

Question

The Tl in a 9.76-g sample of rodenticide was oxidized to the trivalent state and treated with an unmeasured excess of Mg/EDTA solution. The reaction is

Tl3+ + MgY2- → TlY- + Mg2+

Titration of the liberated Mg2+ required 13.34 mL of 0.03560 M EDTA. Calculate the % of Tl2SO4 (504.8 g/mol) in the sample.

Answer 1

The mass $m$ of thallium sulphate can be calculated from its number of the moles $n$ and its molar mass $M$ (504.8 g/mol):

$m = n·M$ .

According to the reaction

Tl³⁺ + MgY^2- → TlY^- + Mg²⁺,

the number of the moles of thallium ions that react equals the number of the moles of magnesium ions liberated:

$n(Tl^{3+}) = n(Mg^{2+})$ .

According to the reaction of titration of the liberated magnesium ions with EDTA,

the number of the moles of the liberated magnesium ions equals the number of the moles of EDTA used:

$n(Mg^{2+}) = n(EDTA) = cV$

$n(Mg^{2+}) = 13.34·10^{-3}·0.03560 = 0.4749·10^{-3}$ mol.

Therefore, the number of the moles of thallium ions is 0.4749·10^-3 mol. Importantly, there are two thallium ions per one molecule of Tl₂SO₄. So, the number of the moles of thallium sulfate is the number of the moles of thallium ions divided by two:

$n(Tl_2SO_4) = \frac{n(Tl^{3+})}{2} = 0.2375·10^{-3}$ mol.

So, the mass of thallium sulphate is:

$m = n·M = 0.2375·10^{-3}·504.8=0.1199$ g.

Finally, the percentage of thallium sulphate in the sample is:

$w, \% = \frac{0.1199}{9.76}·100\% = 1.23\%$ .

Answer: There is 1.23% of Tl₂SO₄ in the sample.