Calculate the pH after 75.0 mL of 0.100 M HO has been added to 100.0 mL of 0.100 M NH₃ (Figure 17.7).

Interpretation:

The value of pH has to be calculated at a point when 100 mL, 0.100 M NH3 is titrated against 75 mL, 0.100 M HCl.

Concept introduction:

Titration is a quantitative method to determine the quantity of an acid or base in a solution. This method is used to determine the concentration an acid in the solution by titrating it against a base. There are four types of acid-base titrations.

(1) Strong acid-Strong base, in this type of titration a strong acid is titrated against a strong base for example, HCl is titrated against NaOH.

(2) Strong acid-Weak base, in this type of titration a strong acid is titrated against a weak base for example, HCl is titrated against NH4OH.

(3) Weak acid-Strong base, in this type of titration a weak acid is titrated against a strong base for example, CH3COOH is titrated against NaOH.

(4) Weak acid-Weak base, in this type of titration a weak acid is titrated against a weak base for example, CH3COOH is titrated against NH4OH.

For weak base-strong acid titration the pH value can be calculated at various points before and after equivalence point.

The equilibrium established during the titration of NH3 with HCl. The equilibrium can be represented as,

NH3(aq)+ H3O+(aq)⇌H2O(l)+NH4+(aq)

Here, the H3O+ ion are generated from HCl, and the reaction is written as,

HCl(aq)+H2O(l)→H3O+(aq)+Cl−(aq)

At equivalence point all the base will be neutralized, and there will be only NH4+ ion and H3O+. The H3O+ will be produced due to the hydrolysis of NH4+ at equivalence point. The hydrolysis equilibrium is represented as,

NH4+(aq)+H2O(l) ⇌H3O+(aq)+NH3(aq)

By using the value of Ka for the NH4+ concentration of H3O+ can be calculated.

Ka=[H3O+][H2O][NH4+] (1)

As H3O+ ions are produced due to hydrolysis of NH4+, the value of pH will be lower than 7 at equivalence point for the weak base-strong acid titrations.

The pH calculation just before the equivalence point is done by using Henderson-Hasselbalch equation.

As the addition of HCl is done there will be formation of buffer solution NH3/NH4+. The pOH calculation for buffer solution is done by using Henderson-Hesselbalch equation.

pOH=pKb+log[conjugate acid][base] (2)

The pH value will be calculated by using expression, pH + pOH = 14.

Given:

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

The value of Kb for ammonia is 1.8×10−5.

The pKb value is calculated as follows;

pKb=−log(Kb)

Substitute, 1.8×10−5 for Kb.

pKb=−log(1.8×10−5)=4.74

Therefore, pKb value is 4.74.

The initial concentration of NH3 is 0.100 mol⋅L−1.

The initial concentration of HCl is 0.100 mol⋅L−1.

The volume of NH3 is 100 mL.

Conversion of 100 mL into L.

(100 mL)(1 L1000 mL)=0.1 L

The volume of HCl is 75 mL

Conversion of 75 mL into L.

(75mL)(1 L1000 mL)=0.075 L

The total volume after the reaction is calculated as,

total volume=volume of NH3(L) + volume of HCl(L)

total volume = 0.1(L)+0.075(L)=0.175 L

Therefore, total volume after reaction is 0.175 L.

The calculation of moles is done by using the expression,

Number of moles=concentration(mol⋅L−1)⋅volume(L)

The ICE table (1) for the reaction between NH3 and HCl is given below,

EquationNH3(aq)+H3O+(aq)⇌H2O(l)+NH4+(aq)Initial(mol)0.01000.00750Change(mol)−0.0075−0.00750.0075After reaction(mol)0.002500.0075

From ICE table (1),

Number of moles of ammonia left after reaction are 0.0025 mol.

Number of moles of H3O+ left after reaction are 0 mol