   Chapter 17, Problem 47PS

Chapter
Section
Textbook Problem

You add 0.979 g of Pb(OH)2 to 1.00 L of pure water at 25 °C. The pH is 9.15. Estimate the value of Ksp for Pb(OH)2.

Interpretation Introduction

Interpretation:

Solubility product constant Ksp value for Pb(OH)2 has to be calculated using the given pH value of the solution.

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.

The expression for the pH is as follows,

pH=-log[H+]

The expression for the ionic product of water is,

Kw=[H+][OH]

Explanation

The solubility product constant Ksp for Pb(OH)2 is calculated below.

Given:

pH of the solution of Pb(OH)2 is 9.15.

The expression for the pH is as follows,

pH=-log[H+]

Rearrange the expression for [H+].

[H+]=10pH

Substitute value of pH in the above expression.

[H+]=10pH=109.15=7.079×1010 M

Calculate the concentration of [OH] using the expression for ionic product of water.

Kw=[H+][OH]

The value of ionic product constant of water, Kw is 1.0×1014.

Substitute 7.079×1010M for [H+] and 1.0×1014 for Kw.

[OH]=Kw[H+]=1.0×10147.079×1010M=1.413×105 M (1)

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

Pb(OH)2(s)Pb2+(aq)+2(OH)1(aq)

The ICE table is as follows,

Equation

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