Question

A sample of a weak acid HA is buffered at pH 5.43. It is found that 71.0 % of the acid is in the HA(aq) form. What is Ka for the acid?

Answer 1

Let us first calculate [H⁺] in buffer from pH value.

pH = - log [H⁺]

[H⁺] = 10^-pH.

[H⁺] = 10^-5.43.

[H⁺] = 3.72 x 10^-6 M.

Given that 71.0 % acid HA is in HA (aq) i.e. unionized form.

i.e. 100 - 71 = 29 % is ionized.

And obviously,

29 % = 3.72 x 10^-6 M [H⁺].

then 71 % [HA] = say 'A' M [HA].

On crossing,

[HA] x 29 = 71 x 3.72 x 10^-6 M.

[HA] = 79 x 3.72 x 10^-6 M /29.

[HA] = 2.9 x 10^-5 M.

A balanced ionization reaction,

HA(aq) ---------> H⁺ (aq) + A^- (aq)

We have,

Ka = [H⁺][A^-] / [HA] ----------- ((1)

We have,

[H⁺] = [A^-] = 3.72 x 10^-6 M.  

[HA] = 2.9x 10^-5 M.

Using these equilibrium concentration in eq.(1)

Ka = (3.72 x 10^-6 )(3.72 x 10^-6 .) / ( 2.9 x 10^-5 M.)

Ka = 4.77 x 10^-7.

is the answer

=======================XXXXXXXXXXXXXXXXXXXX=======================