Question

A solution contains 0.45 M hydrofluoric acid (HF; KA = 6.8 X 10−4). Write the dissociation reaction. Determine the degree of ionization and the pH of the solution

Answer 1

Degree of ionization = 0.0377

pH of the solution = 1.769

Step-by-step explanation:

Initial concentration of HF = 0.45 M

`K_a = 6.8 * 10^(-4)`

`HF \leftrightharpoons  H^+ + F^-`

Initial 0.45 0 0

At equi 0.45 - x x x

Equilibrium constant =
`([H^+][F^-])/(HF)`

`6.8 * 10^(-4)= ([x][x])/(0.45 - x)`

`x^2 + 6.8 * 10^(-4) x -  6.8 * 10^(-4) * 4.5 = 0`

x = 0.017 M

x = Cα

α = Degree of ionization

C = Concentration

Degree of ionization =
`(0.017)/(0.45) = 0.0377`

`pH = -log[H^+]`

[H^+]=0.017 M

`pH = -log[0.017]`

= 1.769