   Chapter 17.4, Problem 17.10CYU

Chapter
Section
Textbook Problem

Calculate the solubility of Ca(OH)2 in moles per liter and grams per liter (Ksp = 5.5 × 10−5).

Interpretation Introduction

Interpretation:

Solubility of salt Ca(OH)2 is to be calculated in moll1 and gl1 units using given Ksp value.

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 derived 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

Solubility of salt Ca(OH)2 in  moll1 and gl1 units is to be calculated as follows,

Given:

The value of solubility product Ksp of Ca(OH)2 is 5.5×10-5.

Ca(OH)2 when dissolved in water dissociates as,

Ca(OH)2(s) Ca2+(aq)+ 2OH1(aq)

The ICE table is as follows,

EquationCa(OH)2Ca2++2OH1Initial (M)00Change (M)+s+2sEquilibrium (M) s2s

Ksp expression for Ca(OH)2 is as follows,

Ksp=[Ca2+][OH1]2

Substitute s for [Ca2+] and 2s for [OH1].

Ksp=s(2s)2

Here,

• Ksp is solubility product constant
• s is solubility

Ksp=4s3s3=Ksp4

Rearrange the above expression

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