Answer to Question #120147 in Chemistry for Jhea

Question

2. 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

Solution:

H₂Y^2- = EDTA

The reactions are:

(1) Tl³⁺ + MgY^2- → TlY^- + Mg²⁺ 

(2) Mg²⁺ + H₂Y^2- → MgY^2- + 2H⁺

According to the (2) equation: n(EDTA) = n(Mg²⁺)

Each mole of EDTA reacts with one mole of Mg²⁺.

n(EDTA) = Molarity of EDTA × Volume of solution

n(EDTA) = (0.03560  M) × (0.01334 L) = 0.0004749 mol

n(Mg²⁺) = n(EDTA) = 0.0004749 mol

According to the (1) equation: n(Mg²⁺) = n(Tl³⁺)

For every one mole of Mg²⁺, there one mole of Tl³⁺

n(Tl³⁺) = n(Mg²⁺) = 0.0004749 mol

(3) Tl₂SO₄ → 2Tl³⁺

According to the (3) equation: n(Tl₂SO₄) = n(Tl³⁺)/2

There are two thallium ions per one molecule of Tl₂SO₄.

n(Tl₂SO₄) = n(Tl³⁺)/2 = (0.0004749 mol) / 2 = 0.00023745 mol

Moles of Tl₂SO₄ = n(Tl₂SO₄) = Mass of Tl₂SO₄ / Molar mass of Tl₂SO₄

The molar mass of Tl₂SO₄ is 504.8 g/mol.

So, the mass of thallium sulphate is:

Mass of Tl₂SO₄ = n(Tl₂SO₄) × M(Tl₂SO₄)

Mass of Tl₂SO₄ = (0.00023745 mol) × (504.8 g/mol) = 0.11986 g

Finally, the percentage of thallium sulphate (Tl₂SO₄) in the sample is:

% of Tl₂SO₄ = m(Tl₂SO₄) / m(sample) = (0.11986 g) / (9.76 g) = 0.01228 or 1.23%

% of Tl₂SO₄ = 1.23 %

Answer: 1.23% of Tl₂SO₄ in the sample.