   Chapter 16, Problem 95CWP

Chapter
Section
Textbook Problem

# The Ksp for PbI2(s) 1.4 × 10−8. Calculate the solubility of PbI2(s) in 0.048 M NaI.

Interpretation Introduction

Interpretation: The solubility of PbI2 in 0.048MNaI is to be calculated.

Concept introduction: Solubility product is defined as the mathematical product of the dissolved ion concentration of a substance raised to the power of its stoichiometric coefficients. When sparingly soluble ionic compound releases ions in the solution, it gives relevant solubility product. The solvent is generally water. Precipitate formation takes place in solution if ionic product is greater than the solubility product.

Explanation

Explanation

To determine: The solubility of PbI2 in 0.048MNaI .

Given

The Ksp of PbI2 is 1.4×108 .

The salt PbI2 dissolves according to the equilibrium given below.

PbI2(s)Pb2+(aq)+2I(aq)

Let x mole/L of PbI2 be dissolved to reach equilibrium.

The concentration of NaI is 0.048M . Therefore the concentration of I will be 0.048M .

Initial concentration(mol/L)[Pb2+]=0[I]=0.048to reachequilibriumxmoles/LdissolvesEquilibriumconcentration(mol/L)[Pb2+]=x[I]=0.048+2x

The solubility product expression is,

Ksp=[Pb2+][I]2

Where,

• Ksp is the solubility product of PbI2

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