Question

Answer the following :

The pH of rainwater collected in a certain region of Maharashtra on a particular day was 5.1. Calculate the H⁺ ion concentration of the rainwater and its percent dissociation.

Answer 1

Given: pH of rainwater = 5.1

To find:

i. H⁺ ion concentration

ii. Percent dissociation

Formula:

i. pH = -log₁₀[H₃O⁺]

ii. Percent dissociation =  α × 100

Calculation: From the formula (i),

pH = -log₁₀[H₃O⁺]

∴ log₁₀[H₃O⁺] = -5.1

= -5 - 0.1 + 1 - 1

= (-5 - 1) + 1 - 0.1

= -6 + 0.9 = `bar(6).9`

∴ [H₃O⁺] = Antilog₁₀[`bar(6).9`]

= 7.943 × 10^-6 M

Considering that the pH of rainwater is due to the dissociation of a monobasic strong acid (HA), we have

\[\ce{HA_{(aq)} + H2O_{(l)} -> H3O^+_{ (aq)} + A^-_{ (aq)}}\]

∴ [H₃O⁺] = α

α = `7.943 xx 10^-6`

From formula (ii),

Percent dissociation = `7.943 xx 10^-6 xx 100 = 7.943 xx 10^-4`

i. H⁺ ion concentration is `7.943 xx 10^-6` M

ii. Percent dissociation is `7.943 xx 10^-4`.