   Chapter 17, Problem 7PS

Chapter
Section
Textbook Problem

What is the pH of the buffer solution that contains 2.2 g of NH4Cl in 250 mL of 0.12 M NH3? Is the final pH lower or higher than the pH of the 0.12 M ammonia solution?

Interpretation Introduction

Interpretation:

pH of a buffer solution that contains 2.2 g ammonium chloride,NH4Cl in 250 ml of 0.12M ammonia,NH3 has to be determined. Also compared the pH of the buffer solution with the pH value of pure ammonia solution.

Concept introduction:

In aqueous solution an base undergoes ionization. The ionization of base is expressed in terms of the equilibrium constant. The quantitative measurement tells about the strength of the base. Higher the value of Kb stronger will be the base. The base dissociation can be represented as following equilibrium,

B(aq)+H2O(l)BH+(aq)+OH(aq)

A weak base undergoes partial dissociation in aqueous solution. The expression for the base dissociation constant Kb is given as,

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

Here,

• [BH+](eq) is the equilibrium concentration of conjugate acid of the base.
• [OH](eq) is the equilibrium concentration of hydroxide ion.
• [B](eq) is the equilibrium concentration of base.

The ICE table (1) gives the relationship between the concentrations of species at equilibrium.

EquationB(aq)+H2O(aq)BH+(aq)+OHInitial(M)c00Change(M)x+x+xEquilibrium(M)cxxx

From the ICE table (1),

[BH+](eq)=x[OH](eq)=x[B](eq)=cx

Substitute x for [BH+](eq), x for [OH](eq) and cx for [B](eq) in equation (1).

The base dissociation constant will be

Kb=(x)(x)(cx)=x2(cx)

Kb=x2(cx) (2)

This table can be modified if one of the ion is already present before the ionization of acid. Then there will be some extent of suppression of dissociation of the weak base. This can be explained on the basis of Le-Chatelier’s principle. According to which reaction will be more on the left side rather than right if one the ion from product side is already present before equilibrium. This suppression of ionization of weak base in presence of strong electrolyte having common ion is called as “Common Ion effect”.

Therefore a modified ICE table is used to give the concentration relationships between ions. For example, if cation BH+ is already present in the solution before equilibrium,

EquationB(aq)+H2O(aq)BH+(aq)+OHInitial(M)cy0Change(M)x+x+xEquilibrium(M)cxy+xx

Here,

• y is the initial concentration of the cation BH+(common ion coming from strong electrolyte) present in the solution before the dissociation of a weak base HA.

From the ICE table (2),

[H3O+](eq)=x[A](eq)=x+y[HA](eq)=cx

Substitute x for [H3O+](eq), x+y for [A](eq) and cx for [HA](eq) in (1).

The expression for the base dissociation constant, Kb, will be given as,

Kb=(x)(x+y)(cx)

There is an assumption for common ion effect, according to which the value of “x” is very small on comparing to the initial concentration of base (B)c”and initial concentration of cation (BH+), “y”. Thus x can be neglected with respect to y and c.

Then Kb can be written as,

Kb=(x)(y)c (3)

The expression for pOH is,

pOH=log[OH] (4)

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

pH+pOH=pKw (5)

The value of pKw is 14.0.

Explanation

The value of pH for the given buffer solution and that of pure ammonia solution is calculated as below.

Given:

Refer to the table 16.2 in the textbook for the value of Kb.

The value of base dissociation constant, Kb, for ammonia is 1.8×105.

The initial concentration of ammonia is 0.12M.

The given mass of ammonium chloride is 2.2 g.

The volume of solution is 250 ml.

Calculation of pH value for buffer solution that contains ammonium chloride and ammonia:

Ammonia undergoes dissociation in aqueous solution and the reaction is given as,

NH3(aq)+H2O(aq)NH4+(aq)+OH(aq)

Ammonium chloride is a strong electrolyte and dissociates to give ammonium ion and chloride ion in aqueous solution.

NH4Cl(aq)+H2O(l)NH4+(aq)+Cl(aq)

In presence of ammonium chloride having common ammonium ion, due to the common ion effect suppression of dissociation of ammonia will occur.

Ammonium chloride is a strong electrolyte, therefore, the concentration of ammonium ion coming from ammonium chloride is equal to the initial concentration of ammonium chloride and is calculated as follows,

[NH4Cl]=w(M)(V)  (6)

Here,

• w is the given mass of ammonium chloride.
• M is the gram molecular mass of ammonium chloride.
• is the volume of the solution in liter.

The volume of the solution is,

V =(250 ml)(1.00 L1000 ml)=0.25 L

Gram molecular mass of ammonium chloride is 52.49 gmol1.

Substitute 1.56g for w, 82.034 gmol1 for M and 1 L for in equation (6).

[NH4Cl]=2.2g(53.49 gmol1)(0.25 L) =0.16 molL1=0.16 M

The ICE table (3) is given as follows,

EquationNH3(aq)+H2O(aq)NH4+(aq)+OH(aq)Initial(M)0.120.160Change(M)x+x+xEquilibrium(M)0.12x0.16+xx

The value of x i.e., the concentration of hydroxide ion can be calculated by using equation (3).

Kb=(x×y)c

Rearrange for x.

x=Kb(c)y

Substitute 1.8×105 for Kb, 0.12M for c and 0.16 M for y.

x=(1.8×105)(0.12)0.16=1.35×105

Therefore the hydroxide ion concentration,[OH], is 1.35×105molL1.

Substitute 1.35×105molL1 for [OH] in equation (4)

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