Question

Calculate the degree of ionization of 0.05M acetic acid if its pKa value is 4.74. How is the degree of dissociation affected when its solution also contains (a) 0.01M (b) 0.1M in HCl ?

Answer 1

c = 0.05 M
pKa = 4.74
pKa = -log(Ka)
Ka = 1.82 × 10 - 5
Ka = ca2 a = $\sqrt{\frac{k_a}{c}}$
a = $\sqrt{\frac{1.82 × 10 ^{- 5}}{5 × 10 ^{- 2}}}$
When HCl is added to the solution, the concentration of H+ ions will increase. Therefore, the equilibrium will shift in the backward direction i.e., dissociation of acetic acid will decrease.
Case I: When 0.01 M HCl is taken.
Let x be the amount of acetic acid dissociated after the addition of HCl.
CH3COOH $\leftrightarrow$ H+ CH3COO- Initial conc
Ka = $\frac{[CH_3COO^-][H^+]}{[CH_3COOH]}$
∴ Ka = $\frac{(0.01)x}{0.05}$
x = $\frac{1.82\times10^{-5}\times 0.05}{0.01}$
x = 1.82 × 10-3
Now, a = Amount of acid dissociate / Amount of acid taken = $\frac{1.82 × 10^{-3} ×0.05}{0.05}$ = 1.82 × 10-3
Case II: When 0.1 M HCl is taken.
Let the amount of acetic acid dissociated in this case be X. As we have done in the first case, the concentrations of various species involved in the reaction are : [CH3COOH] = 0.05 - X : 0.05M
[CH3COO-] = X
[H+] = 0.1 + X : 0.1M
Ka = $\frac{[CH_3COO^-][H^+]}{[CH_3COOH]}$
∴ Ka = $\frac{(0.1)K}{0.05}$
x = $\frac{1.82\times10^{-5}\times 0.05}{0.1}$ = x = 1.82 × 10-4 × 0.05M
Now, a = Amount of acid dissociated / Amount of acid taken = $\frac{1.82 × 10^{-4} ×0.05}{0.05}$ = 1.82 × 10-4