   Chapter 14, Problem 125E

Chapter
Section
Textbook Problem

# A 0.050-M solution of the salt NaB has a pH of 9.00. Calculate the pH of a 0.010-M solution of HB.

Interpretation Introduction

Interpretation: A 0.050M solution of the salt NaB is given to have a pH value of 9.00 . The pH of a 0.010M solution of HB is to be calculated.

Concept introduction: The pH of a solution is define as a figure that expresses the acidity of the alkalinity of a given solution.

The pH of a solution is calculated by the formula, pH=log[H+]

The sum, pH+pOH=14

The equilibrium constant for water is denoted by Kw and is expressed as,

Kw=[H+][OH]

The value of Kw is calculated by the formula,

Kw=KaKb

Explanation

Explanation

To determine: The pH of a 0.010M solution of HB .

The pOH of the given NaB salt is 5_ .

The given value of pH is 9.00 .

The sum, pH+pOH=14

The value of pOH is calculated by the formula,

pOH=14pH

Substitute the value of pH in the above expression.

pOH=149=5_

The [OH] is 1.0×10-5M_ .

The pOH of a solution is calculated by the formula,

pOH=log[OH]

Rearrange the above expression to calculate the value of [OH] .

[OH]=10pOH

Substitute the value of pOH in the above expression.

[OH]=105=1.0×10-5M_

The equilibrium constant expression for the given reaction is, Kb=[BH][OH][B]

The dominant equilibrium reaction is,

B(aq)BH(aq)+OH(aq)

At equilibrium, the equilibrium constant expression is expressed by the formula,

Kb=ConcentrationofproductsConcentrationofreactants

Where,

• Kb is the base ionization constant.

The equilibrium constant expression for the given reaction is,

Kb=[BH][OH][B] (1)

The Kb value is 2.0×10-9_ .

The change in concentration of B is assumed to be x .

The liquid components do not affect the value of the rate constant.

The ICE table for the stated reaction is,

B(aq)BH(aq)+OH(aq)Inititialconcentration0.05000Changex+x+xEquilibriumconcentration0.050xxx

The equilibrium concentration of [B] is (0.050x)M .

The equilibrium concentration of [BH] is xM .

The equilibrium concentration of [OH] is xM .

Substitute the value of [B] , [BH] and [OH] in equation (1).

Kb=[x][x][0.050x]Kb=[x]2[0.050x]

The value of [OH] that is equal to the value of x is 1.0×105M .

Substitute the value of x in the above expression.

Kb=[1.0×105]2[0.050(1.0×105)]Kb=2.0×10-9_

The Ka value is 5.0×10-6_

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