Question

The Ka of a monoprotic weak acid is 0.00732. What is the percent ionization of a 0.123 M solution of this acid?

Percent ionization: %

Answer 1

21.60% is the percent ionization of a 0.123 M solution of this acid.

Step-by-step explanation:

The equilibrium reaction for dissociation of weak acidis,

`HA\rightleftharpoons A^-+H^+`

initially conc. c 0 0

At eqm.
`c(1-\alpha )`
`c\alpha`
`c\alpha`

Concentration of the weak acid (c) = 0.123 M

Acid dissociation constant =
`k_a=0.00732`

Degree of ionization of weak acid =
`\alpha`

An expression of dissociation constant is given as:

`k_a=((c\alpha)(c\alpha))/(c(1-\alpha ))=(c* (\alpha )^2)/((1-\alpha ))`

`0.00732=(0.123 M* (\alpha )^2)/((1-\alpha ))`

`\alpha =0.2160`

Percent ionization of weak acid:

`\% ionization=(c\alpha )/(c)* 100`

`(0.123 M* 0.2160)/(0.123 M)* 100=21.60\%`