HELP!! Chemistry problem finding Ka?

Question

HZ is a weak acid. An aqueous solution nominally 0.020M in HZ has a pH of 4.93 at 25 degrees Celsius. What is the Ka for HZ.
Please show all work so I can understand.

Answer 1

HZ --> H+ + Z- This is the reaction

[HZ] --->[H+] + [Z-]

This relates the concentrations of the components: one molecule of HZ forms one molecule each of H+ and Z-. Let the concentrations of H+ = [H+] = x. Then [Z-] also = x. For each H+ or Z- on the right, there is one less molecule of HZ, so the concentration of HZ = [HZ] is reduced by x, or

[HZ] = C-x, where C is the initial concentration (0.02M).

By the definition of Ka,

Ka = [H+]*[Z-] / [HZ] = x*x / C-z = x^2 / C-x.

You were given the pH. The definition of pH = -log[H+]. That means you can calculate [H+] = 10^-4.93 = 1.175*10^-5. This is the value of x, so plug it in:

Ka = 1.175^2*10^-10 / (0.02 - 1.175*10^-5) = 6.906*10^-9

Answer 2

Its debatable and there are actually several possible answers to this question..