3 years ago

If the Ka of a monoprotic weak acid is 5.4 × 10-6, what is the pH of a 0.25 M solution of this acid?

1 answer:

6 0

The first step is to write the balanced equation of the dissociation of the monoprotic weak acid. This can be assumed to have 1:1 stoichiometric relationships, so the equation and ICE chart are given by:

      HA ---> H^+ + A^-
initial    0.25 0 0
change       -x x x
equilibrium    0.25-x x x

This gives an expression for Ka given by:

Ka = (x)^2/(0.25-x) = 5.4x10^-6
x = 1.159x10^-3 = [H+] = [A-]

pH = -log[H+] = -log[1.159x10^-3] = 2.936

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