Question

A 0.285 M solution of the sodium salt, NaA, of the weak monoprotic acid, HA, has a pH of 9.65. Calculate Ka for the acid HA.

I think I am having issues with the equation and then the math part. I figured out what 'x' was from the pH, which I believe is 2.2x10^-10, but when I tried to put it into the rest of the math equation I did not get the answer (I think it is supposed to be 1.43x10^-6).

Can someone walk me through this? Thanks.

-B

Answer 1

1.43 x 10^-6 ix correct.
NaA(aq) ==> Na^+ + A^-
The anion hydrolyzes in water solution to produce a pH of 9.65
A^- + HOH ==> HA + OH^-
Kb = Kw/Ka = (HA)(OH^-)/(A^-)
You know pH. Convert that to pOH and convert that to OH^-. That = (HA) also. You know (A^-) from the problem. That leaves you with one unknown of Ka. Solve for that. I get 1.4284 x 10^-6 which rounds to 1.43 x 10^-6 to three significant figures.

Answer 2

THANKS! I know what I was doing wrong! I didn't find the pOH I just did 10^-9.65 trying to find x! I understand now! Thanks so much!

-B

Answer 3

right. You found (H^+).