   # Calculate the solubility of silver bromide, AgBr, in moles per liter, in pure water. Compare this value with the molar solubility of AgBr in 225 mL of water to which 0.15 g of NaBr has been added. ### 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 58PS
Textbook Problem
677 views

## Calculate the solubility of silver bromide, AgBr, in moles per liter, in pure water. Compare this value with the molar solubility of AgBr in 225 mL of water to which 0.15 g of NaBr has been added.

Interpretation Introduction

Interpretation:

The solubility of silver bromide,AgBr in mole per liter unit, in pure water has to be calculated and also this value with the molar solubility of AgBr in 225 ml of water containing 0.15 gNaBr has to be compared.

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 solubility of a salt decreases if in the solution one of the ion 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, if anion Bx is already present in the solution before equilibrium, therefore a modified ICE table (2) 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 (2),

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

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

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 (2)

### Explanation of Solution

• The molar solubility of the salt AgBr in pure water and in water containing 0.15 gNaBr 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 AgBr is 5.4×1013.

The volume of the solution is 225 ml.

Amount of NaBr added to the solution is 0.15 g.

Calculation of molar solubility of AgBr in pure water:

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

AgBr(s) Ag+(aq)+ Br(aq)

The ICE table(2) is as follows,

EquationAgBr(s)Ag+(aq)+Br(aq)Initial (M)00Change (M)+s+sEquilibrium (M) ss

The Ksp expression for AgBr is,

Ksp=[Ag+][Br1] (3)

From the table,

[Ag+]=s[Br1]=s

Substitute s for [Ag+] and [Br1] in equation (3).

Ksp=(s)(s)

Here,

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

Ksp=s2

Rearrange for s.

s=Ksp2

Substitute 5.4×1013 for Ksp.

s=5.4×10132= 7.35×107 molL1

• Calculation of molar solubility of AgBr in 225 ml of water containing 0.15 gNaBr.

NaBr is a strong salt and undergoes complete dissociation to give sodium ions and bromide ions in the solution and therefore the concentration of both the sodium and bromide ions is equal to the initial concentration of NaBr.

NaBr(s) Na+(aq)+Br(aq)

The concentration of sodium bromide (NaBr) is calculated as follows,

[NaBr]=wMV  (4)

Here,

• w is the given mass of sodium bromide.
• M is the gram molecular mass of the sodium bromide.
• is the volume of the solution in liter.

The volume of the solution is,

V =(225 ml)(1.00 L1000 ml)=0.225L

Gram molecular mass of sodium bromide is 102.9 gmol.

Substitute 0.15 g for w, 102.9 gmol for M and 0.225 L for in equation (4).

[NaBr]=0

### Still sussing out bartleby?

Check out a sample textbook solution.

See a sample solution

#### The Solution to Your Study Problems

Bartleby provides explanations to thousands of textbook problems written by our experts, many with advanced degrees!

Get Started

Find more solutions based on key concepts
For people with diabetes, the risk of heart disease, stroke, and dying on any particular day is cut in half. T ...

Nutrition: Concepts and Controversies - Standalone book (MindTap Course List)

What are the strengths and weaknesses of vegetarian diets?

Understanding Nutrition (MindTap Course List)

What is motion?

An Introduction to Physical Science

How do a solute, a solvent, and a solution differ?

Biology: The Dynamic Science (MindTap Course List)

What does the spectrum of a prominence reveal? What does its shape reveal?

Horizons: Exploring the Universe (MindTap Course List)

You are reshelving books in a library. You lift a book from the floor to the top shell. The kinetic energy of t...

Physics for Scientists and Engineers, Technology Update (No access codes included)

If the holdfast of this alga breaks free in a storm, will the organism die?

Oceanography: An Invitation To Marine Science, Loose-leaf Versin 