   Chapter 17, Problem 43PS

Chapter
Section
Textbook Problem

When 1.55 g of solid thallium(I) bromide is added to 1.00 L of water, the salt dissolves to a small extent. T l B r ( s ) ⇄ T l + ( a q ) + B r − ( a q ) The thallium(I) and bromide ions in equilibrium with TlBr each have a concentration of 1.9 × 10−3 M. What is the value of Ksp for TlBr?

Interpretation Introduction

Interpretation:

Value of solubility product constant Ksp for TlBr has to be calculated.

Concept introduction:

The solubility of a salt is defined as the maximium 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.

Expression for Ksp of a salt is given as,

AxBy(s)xAy+(aq)+yBx(aq)Ksp=[Ay+]x[Bx]y

Relation between Ksp and s is as follows,

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

Ksp=[Ay+]x[Bx]yKsp=(xs)x(ys)y=xxyy(s)x+y

Rearrange the expression for s.

(s)x+y=Kspxxyy=(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

Ksp is calculated by using molar solubility of the salt.

Explanation

The solubility product constant Ksp for TlBr is calculated below.

Given:

The equilibrium concentration, [Tl+1]=[Br1]=1.9×103M.

When is TlBr dissolved in water, it dissociates as follows,

TlBr(s) Tl+1(aq)+ Br1(aq)

Ksp expression for TlBr is given as,

Ksp=[Tl+1][Br1]

Substitute equilibrium concentration of Tl+1 and Br1

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