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

Given: pH of rainwater = 5.1

To find:

i. \[H^+\] ion concentration

ii. Percent dissociation

Formula:

i. \(\text{pH} = -\log_{10}[H_3O^+]\)

ii. Percent dissociation = \(\alpha \times 100\)

Calculation: From formula (i),

\(\text{pH} = -\log_{10}[H_3O^+]\)

Therefore, \(\log_{10}[H_3O^+] = -5.1\)

\(= -5 - 0.1 + 1 - 1\)

\(= (-5 - 1) + 1 - 0.1\)

\(= -6 + 0.9 = \overline{6.9}\)

Therefore, \([H_3O^+] = \text{Antilog}_{10}[\overline{6.9}]\)

\(= 7.943 \times 10^{-6} \, \text{M}\)

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

\[HA_{(aq)} + H_2O_{(l)} \rightarrow H_3O^+_{(aq)} + A^-_{(aq)}\]

Therefore, \([H_3O^+]

=

\alpha\)

\(\alpha = 7.943 \times 10^{-6}\)

From formula (ii),

Percent dissociation \(= 7.943 \times 10^{-6} \times 100 = 7.943 \times 10^{-4}\)

i. \(H^+\) ion concentration is \(7.943 \times 10^{-6}\, \text{M}\)

ii. Percent dissociation is \(7.943 \times 10^{-4}\).