Answer to Question #125939 in Chemistry for Albert Eshun

Question

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

a) 2.5 × 10–5

b) 5.0 × 10–5

c) 4.7 × 10–3

d) None of the above

Answer 1

H₂O_(l) + HAcO_(aq) $\iff$ H3O⁺_(aq) + AcO_(aq)

I 0.85M 0M 0M

C -xM +xM +xM

E (0.85-x) xM xM

k_a = x²/ (0.85-x)

x = [H3O⁺] = 10^-pH-

x = 10^-2.4

x = 0.00398

k_a = (0.00398)²/0.85

k_a = 1.9 x 10^-5

The value of the acid ionization constant, K_a, for acetic acid is 1.9 x 10^-5