The equilibrium concentrations of the reactants and products are [ HA ] = 0.260 M [HA]=0.260 M, [ H + ] = 2.00 * 10 - 4 M [H+]=2.00*10-4 M, and [ A - ] = 2.00 * 10 - 4 M [A-]=2.00*10-4 M. Calculate the value of p K a pKa for the acid HA HA.

pKa of the acid HA with given equilibrium concentrations is 6.8

Explanation:

The dissolution reaction is:

HA ⇔ H⁺ + A⁻

So at equilibrium, Ka is calculated as below

Ka = [H⁺] x [A⁻] / [HA] = 2.00 x 10⁻⁴ x 2.00 x 10⁻⁴ / 0.260

= 15.38 x 10⁻⁸

Hence, by definition,

pKa = - log (Ka) = - log (15.38 x 10⁻⁸) = 6.813