Chapter 17, Problem 62PS

Chemistry & Chemical Reactivity

10th Edition
John C. Kotz + 3 others
ISBN: 9781337399074

Chapter
Section

Chemistry & Chemical Reactivity

10th Edition
John C. Kotz + 3 others
ISBN: 9781337399074
Textbook Problem

Calculate the solubility, in moles per liter, of calcium hydroxide, Ca(OH)2, in a solution buffered to a pH of 12.60.

Interpretation Introduction

Interpretation:

Molar solubility of calcium hydroxide in a solution buffered to a pH12.6 has to be calculated.

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.

The expression for Ksp of a salt is given as,

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

The 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.

The pOH of the solution is calculated by using the relation,

pH+pOH=pKw (1)

The value of pKw is 14.0.

The concentration of [OH] is calculated using the following expression.

[OH]=10pOH (2)

Explanation

The concentration of the Ca2+ ions in the solution gives the true value of molar solubility of Ca(OH)2 not the concentration of the hydroxide ions because the concentration of hydroxide will be affected by the pH of the solution or hydroxide ion already present in the buffer solution as a result of common ion effect. The molar solubility for Ca(OH)2 is calculated as below.

Given:

Refer to the table 17.2 in the textbook for the value of Ksp.

The value of solubility product constant, Ksp, for Ca(OH)2 is 5.5Ã—10âˆ’5.

pH of the buffered solution is 12.6.

Calculate pOH using equation (1).

pH+pOH=pKw

Rearrange the expression for pOH.

pOH=pKwâˆ’pH=Â 14.0âˆ’12.6=1.4

Substitute 1.4 for pOH in equation (2) to calculate concentration of hydroxide ion.

[OHâˆ’]=10âˆ’1.4=0.0398Â M

The initial concentration of hydroxide ion is equal to 0.0398Â M coming from buffer solution.

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

â€‚Â Ca(OH)2(s)â‡ŒCa2+(aq)+2OHâˆ’1(aq)

The ICE table is as follows,

EquationCa(OH)2(s)â‡ŒCa2+(aq)+2OHâˆ’1(aq)InitialÂ (M)00.0398ChangeÂ (M)+s+2sEquilibriumÂ (M)Â sâ€‰0

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