113. 0.02 molar solution of acetic acid is 3 % ionised. Its ionisation constant is a) 0.9 x 10-3 b) 1.8 x 10-5c) 18 x 10-5 d) 1.8 x 10-4
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113. 0.02 molar solution of acetic acid is 3 % ionised. Its ionisation constant is a) 0.9 x 10-3 b) 1.8 x 10-5c) 18 x 10-5 d) 1.8 x 10-4
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Given:
concentration of acetic acid = 0.02M
Solution of acetic acid is 3% ionized
To find:
Ionisation constant
Solution:
The required chemical reaction is-
CH₃COOH + H₂O → H₃O⁺ + CH₃COO⁻
at t=0, C 0 0
at t=t' C-Cα Cα Cα
as we known,
[tex]K_a = \frac{[H_3O^+] [ CH_3COO^-]}{[CH_3COOH]}[/tex]
[tex]K_a = \frac{C\alpha C\alpha }{C-C\alpha }[/tex]
[tex]K_a = \frac{C^2\alpha^2}{C(1- \alpha)}[/tex]
[tex]K_a = \frac{c \alpha ^2}{(1-\alpha )}[/tex]
here, [tex]K_a[/tex] = Ionisation constant
C = concentration
α = ionized
It is given, C = 0.02M
and α = 3% = 0.03
so, putting all the values, we get
[tex]K_a = \frac{(0.02) \times (0.03)^2}{(1-0.03)}[/tex]
[tex]K_a = \frac{0.000018}{0.97}[/tex]
[tex]K_a = 1.86 \times 10^-^5[/tex]
Therefore, Ionisation constant of a solution of acetic acid is found to be 1.86 × 10⁻⁵.