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:
<em>Moles EDTA = Moles Pb²⁺ and Cd²⁺:</em>
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⁺ <em>(1)</em>
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:
<em>Cd²⁺ -Molar mass: 112.411g/mol-:</em>
3.413x10⁻³ moles Cd²⁺ * (112.411g / mol) =
0.384g of Cd²⁺
<em>Pb²⁺ -Molar mass: 207.2g/mol-:</em>
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⁺
