Question

Question 7.47 It has been found that the pH of a 0.01M solution of an organic acid is 4.15. Calculate the concentration of the anion, the ionization constant of the acid and its pKa.

Class XI Equilibrium Page 229

Answer 1

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Answer 2

concept :- organic acids are weak so, dissociate as HA <=> H+ + A-
so, Ionisation constant , Ka = [H+][A-]/[HA]
you can see that ,
for finding Ka we have to require [H+] and [A-]
[H+] , we can find by using the Arrhenius formula, pH = -log[H+]
and [A-], we can find by using the equation
HA <=> H+ + A-
then, calculate Ka and find pKa by using formula pKa = - logKa


HA <=> H+ + A-
pH = -log[H+]
4.15 = -log[H+] => log[H+] = -4.15
$log[H+] = \overline {5}.85$
[H+] = 7.079 × 10^-5 M

hence, [A-] = 7.079 × 10^-5 M
Ka = [ H+][A-]/[HA]
= (7.079 × 10^-5)(7.079 × 10^-5)/0.01
= 5011.22 × 10^-10
= 5.01122 × 10^-7
= 5 × 10^-7 M

pKa = -logKa
= - log(5 × 10^-7)
= 7 - log5 = 7 - 0.699 = 6.301
hence, pKa = 6.301