Calculate the percent dissociation for a 0.015 mol/L acid HA(aq), that has a PH of 3.00.?

Answer 1

If x mol/L of HA dissociates then the concentrations of the particles in the solution are:

[HA] = 0.015 - x M, [A-] = [H3O+] = x M

according to the chem. equation in equilibrium:

HA + H2O <==> A- + H3O+

But we can find [H3O+] from the pH:

pH = 3, [H3O+] = x = 10^(-3)

So, the percent dissociation of HA is:

a = 100*x/C = 100*10^(-3)/0.015 = 0.1/0.015 = 6.67% approx.