For each of the following cases, decide whether the pH is less than 7, equal to 7, or greater than 7.

(a) Equal volumes of 0.10 M acetic acid, CH₃CO₂H, and 0.10 M KOH are mixed.

(b) 25 mL of 0.015 M NH₃ is mixed with 12 mL of 0.015 M HCl.

(c) 150 mL of 0.20 M HNO₃, is mixed with 75 mL of 0.40 M NaOH.

(d) 25 mL of 0.45 M H₂SO₄ is mixed with 25 mL of 0.90 M NaOH.

Interpretation:

The value of pH has to be calculated when equal volume of CH3COOH(0.10 mol⋅L−1) is titrated with NaOH(0.10 mol⋅L−1).

Concept introduction:

For weak acid-strong base titration the pH value can be calculated at various points before and after equivalence point. The equilibrium established during the titration of CH3COOH with NaOH. The equilibrium can be represented as,

CH3COOH(aq)+NaOH(aq)⇌H2O(l)+CH3COONa(aq)

The pH calculation at equivalence point is given below.

Given:

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

The value of Ka for acetic acid is 1.8×10−5.

The value of Kw for water is 1.0×10−14.

The pKa value is calculated as follows;

pKa=−log(Ka)

Substitute, 1.8×10−5 for Ka.

pKa=−log(1.8×10−5)=4.74

Therefore, pKa values is 4.74.

The initial concentration of CH3COOH is 0.100 mol⋅L−1.

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

The volume of NaOH added is equal to the volume of CH3COOH.

Therefore, let the volume of CH3COOH is 100 mL.

Thus volume of NaOH added  will be 100 mL.

total volume=volume of CH3COOH(L) + volume of NaOH(L)

total volume = 0.1(L)+0.1(L)=0.2 L

Therefore, total volume after reaction is 0.2 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 NaOH and CH3COOH is given below,

EquationCH3COOH(aq)+NaOH(aq)⇌H2O(l)+CH3COONa(aq)Initial(mol)0.01000.01000Change(mol)−0.0100−0.01000.0100After reaction(mol)000.0100

From ICE table (1),

Calculate the concentration of acetate ion after reaction.

Substitute, 0.0100 mol for Number of moles and 0.20 L for volume in equation (4).

concentration = 0.01000.20 (mol⋅L−1)=0.05 mol⋅L−1

The concentration of acetate ion after reaction is 0.05 mol⋅L−1.

The acetate ion produced will undergo hydrolysis in presence of water and the reaction equilibrium is written as,

CH3COO−(aq)+H2O(l)⇌OH−(aq)+CH3COOH(aq)

The hydrolysis equilibrium is represented in ICE table (2).

EquationCH3COO−(aq)+H2O(l)⇌OH−(aq)+CH3COOH(aq)Initial(mol⋅L−1)0.050000Change(mol⋅L−1)−x+x+xAfter reaction(mol⋅L−1)0.0500−x+x+x

From ICE table (2),

There is an approximation, that the value of x is very small as comparison to 0

Interpretation:

The value of pH when 25 mL,NH3(0.015 mol⋅L−1) is titrated with 12 mL,HCl(0.015 mol⋅L−1) has to be calculated.

Concept introduction:

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)

Interpretation:

The value of pH when 150 mL,HNO3(0.20 mol⋅L−1) is titrated with 75 mL,NaOH(0.40 mol⋅L−1) has to be calculated.

Concept introduction:

Strong acid-Strong base, in this type of titration a strong acid is titrated against a strong base for example, HNO3 is titrated against NaOH. The reaction between HNO3 and NaOH can be written as follows.

HNO3(aq)+NaOH(aq)→NaNO3(aq)+H2O(l)

Interpretation:

The value of pH when 25 mL,H2SO4(0.45 mol⋅L−1) is titrated with 25 mL,NaOH(0.90 mol⋅L−1) has to be calculated.

Concept introduction:

Strong acid-Strong base, in this type of titration a strong acid is titrated against a strong base for example, H2SO4 is titrated against NaOH. The reaction between H2SO4 and NaOH can be written as follows.

H2SO4(aq)+2NaOH(aq)→Na2SO4(aq)+2H2O(l)

The sulphuric acid is a dibasic acid as it has two hydrogen atoms which can be donated.

Therefore two equivalents of NaOH are required to balance the reaction.