Answer to Question #211986 in Chemistry for cayyy

Question

The Zn in a 0.7457 g sample of foot powder was titrated with 22.57 mL of 0.011639 M EDTA. Calculate the percent of Zn in sample

Answer 1

Solution:

EDTA = Na₂H₂Y×2H₂O = H₂Y^2-

Zn → Zn²⁺

Complexes of EDTA and metal ions (1:1). Thus:

Zn²⁺ + H₂Y^2- → ZnH₂Y

According to the equations above: n(Zn) = n(Zn²⁺) = n(H₂Y^2-)

n(H₂Y^2-) = V(EDTA) × C_M(EDTA) = (22.57 mL) × (1 L / 1000 mL) × (0.011639 mol/L) = 0.0002627 mol

Therefore,

n(Zn) = n(Zn²⁺) = n(H₂Y^2-) = 0.0002627 mol

The molar mass of Zn is 65.38 g/mol.

Therefore,

(0.0002627 mol Zn) × (65.38 g Zn / 1 mol Zn) = 0.017175 g Zn

%Zn = (mass of Zn / mass of sample) × 100%

%Zn = (0.017175 g / 0.7457 g) × 100% = 2.303%

%Zn = 2.303%

Answer: %Zn = 2.303%