   Chapter 16, Problem 25E

Chapter
Section
Textbook Problem

# The concentration of Pb2+ in a solution saturated with PbBr2(s) is 2.14 × 10−2 M. Calculate Ksp for PbBr2.

Interpretation Introduction

Interpretation: The concentration of Pb2+ in a saturated solution of PbBr2 is given. By using this value, the solubility product of PbBr2 is to be calculated.

Concept introduction: 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

Explanation

Explanation

To determine: The solubility product of PbBr2 .

The solubility product of PbBr2 is 3.92×105_ .

Given

Concentration of Pb2+ is 2.14×102M .

Since, solid PbBr2 is placed in contact with water. Therefore, compound present before the reaction is PbBr2 and H2O . The dissociation reaction of PbBr2 is,

PbBr2Pb2+(aq)+2Br(aq)

It is assumed that solubility of ionic solid is s . The solubility is the maximum concentration of salt that can be dissolved in solution.

Make the ICE table for the dissociation reaction of PbBr2 .

PbBr2(s)Pb2+(aq)2Br(aq)Initial(M):00Chang(M)<

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