In a 5.80E-2 M solution of a monoprotic acid HA, the acid is 23.9% dissociated. Calculate Ka for this acid.

Question

jfraser · Answer

the percent dissociation tells you how much of the original acid is really in the dissociated H+ and A- forms:$$HA(aq) \rightleftharpoons H^{+1}(aq) + A^{-1}(aq)$$
since the starting concentration of acid is given, the equilibrium concentration must be the starting concentration MINUS the percentage that's dissociated. Plug into the formula for an acid dissociation equilibrium$$K_A = \frac{[H^{+1}][A^{-1}]}{[HA]}$$

anonymous · Answer

could you use the numbers?

jfraser · Answer

I can, but I think you should try first

anonymous · Answer

I keep getting the wrong answer

jfraser · Answer

go through each step one at a time. What's the [H+] if 23.9% of the original 0.058M solution is dissociated?