Answer to Question #125998 in General Chemistry for Linson

Question

The acidic ingredient in vinegar is acetic acid. The pH of vinegar is around 2.4, and the molar concentration of acetic acid in vinegar is around 0.85 M. Based on this information, determine the value of the acid ionization constant, Ka, for acetic acid
2.5 × 10–5
5.0 × 10–5
4.7 × 10–3
None of the above

Answer 1

Answer

Let, degree of ionization of acetic acid = a

The ionization equation of acetic acid is

CH₃CO₂H         →            CH₃CO₂^- + H⁺

Concentration of acetic acid is 0.85 M

pH = 2.4

Concentration of H⁺  = 10^(-2.4) = 3.98 x 10^-3

Degree of ionization = (3.98 x 10^-3)/0.85 = 4.68 x 10^-3 M

The relation between ionization constant (k_a) , concentration of acid (c) and degree of ionization (a) of a weak acid is

k_a  = ca²

Putting the values of concentration of acid (c) and degree of ionization (a) of acetic acid, we get

k_a  = 1.9 x 10^-5

So, none of the options are correct.