Question

at 0.01M solution of acetic acid is 1.34%ionised at 298k what is ionisation constant of acetic acid​

Answer 1

aq) C Ans. HA Initial Conc. conc. atC .

Answer 2

Ionisation constant k= 1.8199×$10^{-6}$ M

Explanation:

The equation for the reaction is

                                         CH₃COOH ⇆ CH₃COO⁻ + H⁺

initial concentrations:            c                     0               0

after dissociation                  c(1-α)                cα            cα

ionisation constant , k= [CH₃COO⁻][H⁺]/ [CH₃COOH]

i.e, k=  (cα)(cα)/c(1-α)

given c= 0.01M

α= 0.0134

k= $\frac{[(0.01)(0.0134)]^{2} }{(0.01)(1-0.0134)}$

k= 1.8199×$10^{-6}$