A0.50 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

c. electrical charges of ionic bonds.

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