   Chapter 16, Problem 44E

Chapter
Section
Textbook Problem

# The solubility of Pb(IO3)(s) in a 0.10-M KIO3 solution is 2.6 × 10−11 mol/L. Calculate Ksp for Pb(IO3)2(s).

Interpretation Introduction

Interpretation: The solubility of Pb(IO3)2 and the concentration of KIO3 solution is given. The solubility product of Pb(IO3)2 is to be calculated in a given concentration of KIO3 solution.

Concept introduction: Solubility is defined as the maximum amount of solute that can dissolve at a certain amount of solvent at certain temperature. The solubility product, Ksp is the equilibrium constant that is applied when salt partially dissolve in a solvent. The solubility product of dissociation of AxBy is calculated as,

Ksp=[A]x[B]y

Where,

• x is coefficient of concentration of A .
• y is coefficient of concentration of B .
Explanation

Explanation

To determine: The solubility product of Pb(IO3)2 in 0.10M KIO3 .

The solubility product of Pb(IO3)2 in 0.10M KIO3 is 2.6×1013_ .

Given

Solubility of Pb(IO3)2 is 2.6×1011mol/L .

Concentration of KIO3 is 0.10M .

The major species in the solution before dissociation of Pb(IO3)2 are K+,IO3 and H2O . The dissociation reaction of Pb(IO3)2 is,

Pb(IO3)2s)Pb2+(aq)+2IO3(aq)

The ratio of moles between ions is 1:2 .

It is assumed that smol/L of solid is dissolved to reach the equilibrium. The meaning of 1:2 stoichiometry of salt is,

smol/LPb(IO3)21smol/LPb2++2smol/LIO3

The concentrations are as follows,

Initialconcentration(mol/L)Equilibriumconcentration(mol/L)[Pb2+]=0[Pb2+]=s[IO3]=0

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