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:

The value of pH has to be calculated for the resulting solution obtained by adding 25 mL0.12 M hydrochloric acid and 25 mL 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.) + CH3COO−1(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.) + OH−1(aq.)

The pH value for the resulting obtained by adding 25 mL0.12 M hydrochloric acid and 25 mL 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 1.8×10−5.

The concentration of hydrochloric acid is 0.12 M.

The volume of hydrochloric acid is,

(25 mL)(1 L1000 mL)=0.025 L

The concentration of ammonia solution is 0.43 M.

The volume of ammonia solution is,

(25 mL)(1 L1000 mL)=0.025 L

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,

Number of moles=(concentration)(volume) (1)

For hydrochloric acid,

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

Number of moles=(0.12 M)(0.025 L)=0.003 mol

For ammonia solution,

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

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

The ICE table (1) is as follows,

EquationHCl(aq)+NH3(aq)⇌NH4+(aq)+Cl−(aq)Initial(mol)0.0030.0107500Change(mol)−0.003−0.003+0.003+0.003After reaction(mol)07.75×10−30.0030.003

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 0.05 L 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,

Number of moles=(concentration)(volume)

Rearrange it for concentration.

concentration = Number of molesvolume (mol⋅L−1) (2)

For ammonia solution,

Substitute 7.75×10−3 for number of moles and 0.05 L for volume in equation (2).

concentration = 7.75×10−3 mol0.05 L=0.155 mol⋅L−1=0.155 M

For ammonium ion,

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

concentration = 0.003 mol0.05 L=0.06 mol⋅L−1=0.06 M

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