# What is the solubility, in milligrams per milliliter, of BaF 2 , (a) in pure water and (b) in water containing 5.0 mg/mL KF?

### Chemistry & Chemical Reactivity

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

Chapter
Section

### Chemistry & Chemical Reactivity

9th Edition
John C. Kotz + 3 others
Publisher: Cengage Learning
ISBN: 9781133949640
Chapter 17, Problem 60PS
Textbook Problem
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## What is the solubility, in milligrams per milliliter, of BaF2, (a) in pure water and (b) in water containing 5.0 mg/mL KF?

(a)

Interpretation Introduction

Interpretation:

The solubility of Barium fluoride BaF2 salt in milligrams per milliliter unit has to be calculated in pure water.

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 (1) 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 the cation A+y.
• y is the coefficient of the anion Bx.
• s is the molar solubility.

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

### Explanation of Solution

The solubility of the salt BaF2 in pure water is calculated below.

Given:

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

The value of solubility product constant,Ksp of BaF2 is 1.8×107.

The balanced chemical reaction for the dissolution of BaF2 in water is,

BaF2(s) Ba2+(aq)+ 2F(aq)

The ICE table(2) is as follows,

EquationBaF2(s)Ba2+(aq)+2F(aq)Initial (M)00Change (M)+s+2sEquilibrium (M) s2s

The Ksp expression for BaF2 is,

Ksp=[Ba2+][F1]2 (2)

From the table,

[Ba2+]=s[F1]=2s

Substitute s for [Ba2+] and 2s for [F1] in equation (2).

Ksp=(s)(2s)2

Here,

• Ksp is solubility product constant
• s is the molar solubility of the salt AgI.

Ksp=4s3

Rearrange for s.

s=Ksp43

Substitute 1.8×107 for Ksp

(b)

Interpretation Introduction

Interpretation:

The solubility of salt BaF2 has to be calculated in water containing 5.0 mgmL1 KF and also compared it with the solubility value in pure water.

Concept introduction:

The solubility of a salt decreases if in the solution one of the ions common to the dissolved ion of salt is already present before the dissolution of salt due to common ion effect. This can be explained on the basis of Le-Chatelier’s principle. According to which reaction will be more on the left side rather than right if one the ion from product side is already present before equilibrium.

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 (4)

For example, if anion Bx is already present in the solution before equilibrium, therefore a modified ICE table (3) is used to give the concentration relationships between ions.

EquationAxBy(s)=xAy+(aq)+yBx(aq)Initial(M)0s'Change(M)+xs+ysEquilibrium(M)xss'+ys

Here,

• s' is the initial concentration of the anion Bx (common ion coming from strong electrolyte) present in the solution before the dissociation of a weak salt.

From the ICE table (3),

[Ay+]=xs[Bx]=s'+ys

Substitute xs for [Ay+] and s'+ys for [Bx] in equation (4).

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

The value of s is very small in comparison to the value of s'. Therefore it can be neglected and the expression of Ksp changes to,

Ksp=(xs)x(s')y

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