Question

After dissolving in 50.0 mL of good water, the zinc in a 0.422 g sample of foot powder was titrated at a pH=12.0 with 26.67 mL of a 0.01222 M EDTA solution. Calculate the % Zn in this sample and the ppm of Zn in the original foot powder sample.

Answer 1

Given;

EDTA solution used is 26.67 mL of 0.01222M for the complete titration;

Now;

Because EDTA and Zn make 1:1 complex so, Molarity = Normality;

Using the formula; M₁V₁ = M₂V₂                where M₁ = Molarity of EDTA, V₁ = vol. of EDTA used

                                                                M₂ = molarity of Zn solution, V₂ = sample vol. of Zn solution

• 0.01222 X 26.67 = M₂ X 50
• M₂ = 0.00651 M (implies 0.00651 moles of Zn in 1000mL of sample solution)

= 0.00651 X 65 (mol. wt. of Zn) g of Zn in 1000mL of sample solution = 0.423g in 1000mL of sample solution ------- eqn1

So, weight of Zn in 50mL (V₂) of sample solution = (0.423X 50)/1000 = 0.02115g

Percentage of Zn in sample = (0.02115 X 100)/0.422 = 5.01% (answer)                     where 0.422g is weight of foot sample (given)

ppm of Zn in the sample solution = weight of Zn (mg) in 1L of sample = 423 ppm (from eqn. 1)