# What is the pH of the solution that results from adding 25.0 mL of 0.12 M HCl to 25.0 mL of 0.43 M NH 3 ?

### Chemistry & Chemical Reactivity

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

Chapter
Section

### Chemistry & Chemical Reactivity

9th Edition
John C. Kotz + 3 others
Publisher: Cengage Learning
ISBN: 9781133949640
Chapter 17, Problem 6PS
Textbook Problem
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## What is the pH of the solution that results from adding 25.0 mL of 0.12 M HCl to 25.0 mL of 0.43 M NH3?

Interpretation Introduction

Interpretation:

The value of pH has to be calculated for the resulting solution obtained by adding 25mL0.12 M hydrochloric acid and 25mL 0.43 M ammonia solution.

Concept introduction:

A buffer solution is defined as a solution which does not show any change in its pH values on the addition of small amount of acid or base. There are two types of buffer solutions,

(1) Acidic Buffer and (2) Basic Buffer

Acidic Buffer is a solution which has the pH value below 7. An acidic buffer is prepared by mixing a weak acid with its conjugate base. For example, a mixture of weak acid CH3COOH and its conjugate base CH3COO gives an acidic buffer.

CH3COOH(aq.)+ H2O(l)H3O+(aq.)+ CH3COO1(aq.)

Basic Buffer is a solution which has the pH value above 7. A basic buffer is prepared by mixing a weak base with its conjugate acid. For example, a mixture of weak base NH4OH and its conjugate acid NH4+ gives a basic buffer.

NH4OH(aq.)NH4+(aq.)+ OH1(aq.)

### Explanation of Solution

The pH value for the resulting obtained by adding 25mL0.12 M hydrochloric acid and 25mL 0.43 M ammonia solution is calculated below.

Given:

Refer to the table no. 16.2 in the textbook for the value of Kb.

The value of Kb for ammonia is 1.8×105.

The concentration of hydrochloric acid is 0.12 M.

The volume of hydrochloric acid is,

(25mL)(1L1000mL)=0.025L

The concentration of ammonia solution is 0.43 M.

The volume of ammonia solution is,

(25mL)(1L1000mL)=0.025L

Hydrochloric acid reacts with ammonia and the reaction involved is,

HCl(aq)+NH3(aq)NH4+(aq)+Cl(aq)

The initial number of moles of reactants hydrochloric acid and ammonia is calculated by using the following expression,

Numberof moles=(concentration)(volume) (1)

For hydrochloric acid,

Substitute 0.12 M for concentration and 0.025L for volume in equation (1).

Numberof moles=(0.12 M)(0.025 L)=0.003mol

For ammonia solution,

Substitute 0.43 M for concentration and 0.025 L for volume for ammonia solution,

No.of moles=(0.43 M)(0.025L)=0.01075mol

The ICE table (1) is as follows,

EquationHCl(aq)+NH3(aq)NH4+(aq)+Cl(aq)Initial(mol)0.0030.0107500Change(mol)0.0030.003+0.003+0.003Afterreaction(mol)07.75×1030.0030.003

After the reaction, ammonia, ammonium ions, and chloride ions will be left in the reaction mixture as hydrochloric acid is limiting reagent. Ammonia and ammonium ion form a basic buffer and the total volume of the solution is equal to 0.05 L.

Therefore the remaining moles of ammonia will be dissolved in 0.05L and it undergoes dissociation as follows,

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

The concentration of ammonia solution and ammonium ion after the reaction is calculated using the following expression,

Numberof moles=(concentration)(volume)

Rearrange it for concentration.

concentration = Numberof molesvolume(molL1) (2)

For ammonia solution,

Substitute 7.75×103 for number of moles and 0.05L for volume in equation (2).

concentration=7.75×103 mol0.05L=0.155molL1=0.155M

For ammonium ion,

Substitute 0.003 for number of moles and 0.05L for volume in equation (2).

concentration=0.003 mol0.05L=0.06molL1=0.06M

There will be equilibrium between ammonia and ammonium ion which will form basic buffer solution. That can be represented in ICE table (2) as follows,

EquationNH3(aq)+H2O(l)NH4+(aq)+OH(aq)Initial(M)0

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