Solved Example Problems with Answers, Solution

Types of Chemical Reactions

PROBLEMS

Example 1: Calculate the pH of 0.001 molar solution of HCl.

Solution: HCl is a strong acid and is completely dissociated in its solutions according to the process:

HCl(aq) → H+(aq) + Cl–(aq)

From this process it is clear that one mole of HCl would give one mole of H + ions. Therefore, the concentration of H+ ions would be equal to that of HCl, i.e., 0.001 molar or 1.0 × 10–3 mol litre–1.

Thus, [H+] = 1 × 10–3 mol litre–1

pH = –log10[H+] = –log1010–3

= –(–3 × log10) = –(3 × 1) = 3

Thus, pH = 3

Example 2: What would be the pH of an aqueous solution of sulphuric acid which is 5 × 10–5 mol litre–1 in concentration.

Solution: Sulphuric acid dissociates in water as:

H2SO4(aq) → 2 H+(aq) + SO42–(aq)

Each mole of sulphuric acid gives two mole of H+ ions in the solution. One litre of H2SO4 solution contains 5 × 10 –5 moles of H2SO4 which would give 2 × 5 × 10–5 = 10 × 10–5 or 1.0 × 10–4 moles of H+ ion in one litre of the solution.

Therefore,

[H+] = 1.0 × 10–4 mol litre–1

pH = –log10[H+] = –log1010–4 = –(–4 × log1010)

= –(–4 × 1) = 4

Example 3: Calculate the pH of 1 × 10–4 molar solution of NaOH.

Solution: NaOH is a strong base and dissociates in its solution as:

NaOH(aq) → Na+(aq) + OH–(aq)

One mole of NaOH would give one mole of OH– ions. Therefore,

[OH–] = 1 × 10–4 mol litre–1

pOH = –log10[OH–] = –log10 × [10–4]

 = –(–4 × log1010)= –(–4) = 4

Since, pH + pOH = 14

pH = 14 – pOH = 14 – 4

= 10

Example 4: Calculate the pH of a solution in which the concentration of the hydrogen ions is 1.0 × 10–8 mol litre–1.

Solution: Here, although the solution is extremely dilute, the concentration given is not of an acid or a base but that of H+ ions. Hence, the pH can be calculated from the relation:

pH = –log10[H+]

given [H+] = 1.0 × 10–8 mol litre–1

pH = –log1010–8 = –(–8 × log1010)

= –(–8 × 1) = 8

Example 5: If the pH of a solution is 4.5, what is its pOH?

Solution:

pH + pOH = 14

pOH = 14 – 4.5 = 9.5

pOH = 9.5