Question

The Cd & Pb ions in a 50.00-mL req’d 40.19 mL of 0.004870 M
EDTA for titration. A 75.00-mL portion of the same sample was made to be
basic & treated with x’ss KCN, masking the Cd as Cd(CN)4
2-. This soln req’d 31.45 mL of the EDTA for titration. Calc. conc. of Cd2+
(112.41 g/mol) & Pb2+ (207.2 g/mol) in the sample in ppm.

Answer 1

When Cd ions are masked, only Pb ions participate in the reaction with EDTA:

Pb²⁺ + HY₃^- + OH^- $\rightarrow$ PbY^2- + H₂O.

The number of the moles of EDTA used for titration equals the number of the moles of Pb ion in the portion titrated. Therefore, the concentration of Pb ion in the portion is:

$c(Pb^{2+}) = \frac{n(Pb^{2+})}{V} = \frac{n(EDTA)}{75.00·10^{-3}\text{ L}}$

$n(EDTA) = c_Y·V_{Y} = 0.004870·31.45·10^{-3}$

$n(EDTA) = 0.1532·10^{-3}$ mol

$c(Pb^{2+}) = \frac{0.1532·10^{-3}}{75.00·10^{-3}} = 0.002042$ M.

When Cd ions are not masked, they participate in titration. The number of the moles of EDTA used for titration is therefore the sum of the number of the moles of lead and the number of the moles of cadmium in the portion:

$n(EDTA) = n(Pb^{2+}) + n(Cd^{2+})$ .

The number of the moles of lead in 50 mL portion is:

$n(Pb^{2+}) = cV = 0.002042·50·10^{-3} = 0.1021$ mmol.

The number of the moles of EDTA used for titration:

$n(EDTA)=cV = 40.19·10^{-3}·0.004870 =0.1957$ mmol.

Therefore, the number of the moles of Cd²⁺ and its concentration are:

$n(Cd^{2+}) = 0.1957-0.1021 = 0.09362$ mmol

$c(Cd^{2+}) = \frac{n}{V} = \frac{0.09362·10^{-3} \text{mol}}{50·10^{-3}\text{ L}} = 0.001872$ M.

For dilute solutions, one part per million equals one mg/L. The ppm of Cd²⁺ and of Pb²⁺ are:

$Cd^{2+} : 0.001872 \text{ mol/L}·112.41\text{g/mol}·10^{3} = 210.5$ ppm

$Pb^{2+} : 0.002042 \text{ mol/L}·207.2\text{g/mol}·10^{3} = 423.1$ ppm.

Answer: Cd²⁺ and Pb²⁺ concentrations are 210.5 ppm and 423.1 ppm, respectively.