   # What mass of ammonium chloride, NH 4 Cl, must be added to exactly 5.00 × 10 2 mL of 0.10 M NH 3 solution to give a solution with a pH of 9.00? ### Chemistry & Chemical Reactivity

9th Edition
John C. Kotz + 3 others
Publisher: Cengage Learning
ISBN: 9781133949640

#### Solutions

Chapter
Section ### Chemistry & Chemical Reactivity

9th Edition
John C. Kotz + 3 others
Publisher: Cengage Learning
ISBN: 9781133949640
Chapter 17, Problem 10PS
Textbook Problem
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## What mass of ammonium chloride, NH4Cl, must be added to exactly 5.00 × 102 mL of 0.10 M NH3 solution to give a solution with a pH of 9.00?

Interpretation Introduction

Interpretation:

Mass of ammonium chloride that must be added to 5.00×102 ml of 0.1 M ammonia solution such that the resulting solution has pH 9.00 is to be determined.

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 pOH of the solution is calculated by using the relation,

pH+pOH=pKw (4)

The value of pKw is 14.0.

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

pOH=log[OH]

Above equation can be rearranged for [OH]. Therefore expression for [OH] is given as,

[OH]=10pOH (5)

### Explanation of Solution

The mass of ammonium chloride that must be added to ammonia solution such that the resulting solution has pH 9.00 is calculated 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 0.10M.

The volume of solution is 5.00×102 ml.

pH of the resulting solution is 9.00.

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.

The expression for the dissociation constant Kb for ammonia

Kb=[NH4+](eq)[OH](eq)[NH3](eq) (6)

Rearrange equation (4) for pOH.

pOH=pKwpH

Substitute 9.0 for pH and 14.0 for pKw.

pOH=14.09.0=5.0

Substitute 5.0 for pOH in equation (5) to calculate the concentration of hydroxide ions in the solution.

[OH]=105.0=1.0×105 (7)

The ICE table (3) is given as follows,

EquationNH3(aq)+H2O(aq)NH4+(aq)+OH(aq)Initial(M)0.10y0Change(M)x+x+xEquilibrium(M)0.10xy+xx

Here,

• x is the number of moles of ammonia dissociated per liter and is equal to the concentration of hydroxide ion, [OH] and ammonium ion, [NH4+] coming from acetic acid in presence of ammonium chloride.
• y is the initial concentration of the ammonium ions present in the solution coming from strong electrolyte ammonium chloride.
• y+x is the total concentration of ammonium ions coming from ammonia as well as ammonium chloride.

From equation (7) and ICE table (3),

x=1.0×105

From the ICE table (3),

[NH3]=0

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