   # Estimate the solubility of lead(II) bromide (a) in moles per liter and (b) in grams per liter of pure water. ### Chemistry & Chemical Reactivity

9th Edition
John C. Kotz + 3 others
Publisher: Cengage Learning
ISBN: 9781133949640

#### Solutions

Chapter
Section ### Chemistry & Chemical Reactivity

9th Edition
John C. Kotz + 3 others
Publisher: Cengage Learning
ISBN: 9781133949640
Chapter 17, Problem 52PS
Textbook Problem
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## Estimate the solubility of lead(II) bromide (a) in moles per liter and (b) in grams per liter of pure water.

a)

Interpretation Introduction

Interpretation:

The solubility of salt PbBr2 in pure water, in moles per liter using the value of Ksp for PbBr2 has to be estimated.

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.

For example general salt AxBy when dissolved in water dissociates as,

AxBy(s)xAy+(aq)+yBx(aq)

The expression for Ksp of a salt is,

Ksp=[Ay+]x[Bx]y (1)

The ICE table for salt AxBy, which relates the equilibrium concentration of ions in the solution is given as follows,

EquationAxByxAy++yBxInitial(M)00Change(M)+xs+ysEquilibrium(M)xsys

From the table,

[Ay+]=xs[Bx]=ys

Substitute xs for [Ay+] and ys for [Bx] in equation (1).

Ksp=(xs)x(ys)y=xxyy(s)x+y

Rearrange for s.

s=(Kspxxyy)1/(x+y)

Here,

• x is the coefficient of cation A+y.
• y is the coefficient of anion Bx.
• s is the molar solubility.

The value of Ksp is calculated by using molar solubility of the salt.

Solubility in grams per liter for a given salt is calculated by multiplying the molar solubility with the molar mass of the salt.

### Explanation of Solution

The solubility of salt PbBr2 in moles per liter unit is calculated below.

Given:

Refer to the Appendix J in the textbook for the value of Ksp.

The value of solubility product Ksp of PbBr2 is 6.6×106.

The balanced chemical reaction  for the dissolution of PbBr2 in water is,

PbBr2(s) Pb2+(aq)+ 2Br1(aq)

The ICE table(1) is as follows,

EquationPbBr2(s)Pb2+(aq)+2Br1(aq)Initial (M)00Change (M)+s+2sEquilibrium (M) s2s

The Ksp expression for PbBr2 is,

Ksp=[Pb2+][Br1

(b)

Interpretation Introduction

Interpretation:

The solubility of salt PbBr2 in pure water, in grams per liter unit using the value of Ksp for PbBr2 has to be estimated.

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.

For example general salt AxBy when dissolved in water dissociates as,

AxBy(s)xAy+(aq)+yBx(aq)

The expression for Ksp of a salt is,

Ksp=[Ay+]x[Bx]y (1)

The ICE table for salt AxBy, which relates the equilibrium concentration of ions in the solution is given as follows,

EquationAxByxAy++yBxInitial(M)00Change(M)+xs+ysEquilibrium(M)xsys

From the table,

[Ay+]=xs[Bx]=ys

Substitute xs for [Ay+] and ys for [Bx] in equation (1).

Ksp=(xs)x(ys)y=xxyy(s)x+y

Rearrange for s.

s=(Kspxxyy)1/(x+y)

Here,

• x is the coefficient of cation A+y.
• y is the coefficient of anion Bx.
• s is the molar solubility.

The value of Ksp is calculated by using molar solubility of the salt.

Solubility in grams per liter for a given salt is calculated by multiplying the molar solubility with the molar mass of the salt.

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