Question

Calculate the solubility in g/l of lead (ii) fluoride in water at 25 oc if ksp of pbf2 is 4. 1 x 10-8.

Answer 1

The solubility in g/l of lead (ii) fluoride in water at 25°C if ksp of PbF2 is 4. 1 x 10-8. is 5.3 ×10^(-1).

When PbF₂ dissolves, it dissociates as follows;

PbF₂ --> Pb²⁺ + 2F⁻

If molar solubility of PbF₂ is x, then molar solubility of Pb²⁺ is x and F⁻ is 2x.

ksp is solubility product constant and it can be calculated as follows;

ksp = [Pb²⁺][F⁻]²

ksp = (x)(2x)²

ksp = 4x³

$4_{X} ^{3} = 4.1 * 10^{-8}\\\\ X^{3} = 1.025 * 10^{-8}\\\\ X = 0.092 * 10^{-3}\\ \\X = 2.1 * 10^{-3}$

To calculate solubility in terms of g/L, we have to know the molar mass of PbF₂.

Solubility of PbF₂ is $2.1 * 10^{-3}$ mol/L x 248 g/mol = $5.14 * 10^{-1}$ g/L

Thus, we concluded that the solubility of given solution is 5.14 × 10^(-1). which is approx 5.3 × 10^(-1).Hence option D is the correct option.

Learn more about solubility: brainly.com/question/14366471

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