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.
4. Answer the following :
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 \(7.943 \times 10^{-4}\).