   Chapter 17, Problem 54PS

Chapter
Section
Textbook Problem

If 55 mg of lead(II) sulfate is placed in 250 mL of pure water, does all of it dissolve? If not, how much dissolves?

Interpretation Introduction

Interpretation:

From Ksp value for PbSO4 whether all 55 mg radium sulphate placed in 250ml of water will dissolve or not has to be predicted. The amount of PbSO4 dissolved also has to be calculated.

Concept introduction:

The solubility of a salt is defined as the maximum amount of salt that can be dissolved in definite amount of solvent. It is expressed in moles per liter or grams per liter. Solubility in terms of moles per liter is called molar solubility and is defined as the number of moles of solute (salt) dissolved in per liter of solution.

Solubility product constant Ksp is an equilibrium constant and is defined as the product of the equilibrium concentration of the ions of the salt raised to the power of their coefficients in the balanced chemical equation.

The expression for Ksp of a salt is given as,

AxBy(s)xAy+(aq)+yBx(aq)Ksp=[Ay+]x[Bx]y

Explanation

Solubility product constant Ksp for PbSO4 is 2.5×104.

Amount of radium sulphate dissolved in 250ml is 55 mg.

PbSO4 Dissociates as follows in water,

PbSO4(s)Pb2+(aq)+SO42(aq)

The expression for Ksp,

Ksp=[Pb2+][SO42] (1)

The ICE table (1) is as follows,

EquationRaSO4(s)Pb2+(aq)+SO42(aq)Initial (M)00Change (M)+s+sEquilibrium (M) ss

Here, s is the molar solubility of PbSO4.

Substitute value of the concentration of lead and sulphate ions in equation (1) from the table

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