Question

The pH of a 100.0 mL 0.125 M solution of HA is measured to be 4.1.  Calculate Ka for this monoprotic acid.

Answer 1

By definition, in equilibrium:

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

Then

Assume that [H+] = [A-] = x … due to stoichiometry (i.e. 1 mol of H+ per each mol of A-)

Then, in equilibrium [HA] = M – x (i.e. the original concentration minus the acid in equilibrium)

Substitute

Ka = (x*x)/(M-x)

Ka = x*x/(0.125-x)

solve for x (quadratic equation)

If we know that

[H+] = 10^-pH = 10^-4.1 = 0.00007943

[H+] = x = 0.00007943 M

and we know that:

[H+] = [A-] = x = 0.00007943

substitute in x

Ka = x*x/(0.125-x)

Ka = (0.00007943)(0.00007943)/(0.125-0.00007943)

Ka = 5.05050*10^-8