Question

Answer the following in one sentence :

Calculate the pH of 0.01 M sulphuric acid.

Answer 1

Given: Concentration of sulphuric acid = 0.01 M

To find: pH

Formula: pH = `-"log"_10["H"_3"O"^+]`

Calculation:

Sulphuric acid (H₂SO₄) is a strong acid. It dissociates almost completely in the water as:

\[\ce{H2SO4_{(aq)} + 2H2O_{(l)} -> 2H3O^+_{ (aq)} +  SO^{2-}_{4(aq)}}\]

Hence, [H₃O⁺] = 2 × c = 2 × 0.01 M = 2 × 10^-2 M

From formula (i),

pH = -log₁₀[H₃O⁺] = -log₁₀[2 × 10^-2] = `-"log"_10"2" - "log"_10"10"^-2`

= `-"log"_10"2" + 2 = 2 - 0.3010`

pH = 1.699

The pH of 0.01 M sulphuric acid is 1.699.