Question

A 1.757-g sample of a / alloy was dissolved in acid and diluted to exactly 250.0 mL in a volumetric flask. A 50.00-mL aliquot of the diluted solution was brought to a pH of 10.0 with an / buffer; the subsequent titration involved both cations and required 28.89 mL of 0.06950 M EDTA. A second 50.00-mL aliquot was brought to a pH of 10.0 with an / buffer, which also served to mask the ; 19.07 mL of the EDTA solution were needed to titrate the . Calculate the percent and in the sample.

Answer 1

Answer:

78.14% Pb²⁺ and 21.86% of Cd²⁺

Explanation:

The first titration involves the reaction of both Pb²⁺ and Cd²⁺

In the second titration, as the buffer is HCN/NaCN, the Cd²⁺ precipitates as Cd(CN)₂ and the only ion that reacts is Pb²⁺

In the first titration:

Moles EDTA = Moles Pb²⁺ and Cd²⁺:

28.89mL = 0.02889L * (0.06950moles / L) = 2.008x10⁻³ moles in the aliquot. In the sample:

2.008x10⁻³ moles * (250.0mL / 50.0mL) =

0.01004 moles = Pb²⁺ + Cd²⁺ (1)

In the second titration:

19.07mL = 0.01907L * (0.06950mol / L) = 1.325x10⁻³ moles Pb²⁺ in the aliquot. In the sample:

1.325x10⁻³ moles Pb²⁺ * (250.0mL / 50.0mL) =

6.626x10⁻³ moles Pb²⁺

That means the moles of Cd²⁺ are:

0.01004 moles = Cd²⁺ + 6.626x10⁻³ moles Cd²⁺

3.413x10⁻³ moles Cd²⁺

The mass of each ion is:

Cd²⁺ -Molar mass: 112.411g/mol-:

3.413x10⁻³ moles Cd²⁺ * (112.411g / mol) =

0.384g of Cd²⁺

Pb²⁺ -Molar mass: 207.2g/mol-:

6.626x10⁻³ moles Pb²⁺ * (207.2g / mol) =

1.373g of Pb²⁺

The percent mass of each ion is:

1.373g Pb²⁺ / 1.757g = 78.14% Pb²⁺

And:

0.384g of Cd²⁺ / 1.757g * 100 = 21.86% of Cd²⁺