Question

The Ka of a monoprotic weak acid is 4.24 × 10-3. What is the percent ionization of a 0.159 M solution of this acid?

Answer #1

For an acid you have

HA + H2O ------>H3O + A-

construct the ICE table

and dont consider water it is a pure solvent

HA + H2O <-------------> H3O+ + A-

Initial: . 0.159 0 0

change: -x +x +x

equilibrium: 0.159-x +x +x

Ka= [H3O+][A-] / [HA]

4.24 x 10^{-3} = [x]^{2}/ [0.159-x]

x2 + 4.24 *10^{-3} x - 0.674 * 10^{-3} = 0

you have to use the quadratic formula to solve for x

I got x= 0.0239

to find the percent ionization of the weak acid:

([H+] at equilibrium / [HA] initial ) x 100 = (0.0239 / 0.159) x
100

= 15.0 %

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