A 0.50 m solution of an acid ha has ph = 2.24. what is the value of ka for the acid?

A 0.50 M solution of a monoprotic acid HA with a pH of 2.24 would be, first, a weak acid, as it does not dissociate fully. This leaves us with an equilibrium expression: HA (aq) ⇌ H + (aq) + A - (aq)

Where A - is the conjugate base of the weak acid.

In a study of equilibrium, we remember that the ka value is the acid dissociation constant, and has the equation:

Ka = (concentration of H+) (concentration of conjugate base) / concentration of acid

We know the concentration of H + and A - are 10^-2.24 by the definition of a pH being the - log (concentration of H+).

The concentration of the acid has gone down a little bit, as it has partially dissociated into H + and A-, so we'll have to subtract 10^-2.24 from 0.50 for the concentration of the acid to account for the dissociation.

The final equation would then become:

[H+]*[A-]/[HA] = Ka

(10^-2.24) * (10^-2.24) / (0.50 - 10^-2.24) = Ka

(3.31 * 10^-5) / (0.494) = Ka

Ka = 6.70 * 10^-5