[bird8584]

the pH of a .0243 M solution of a monoprotic weak acid is 2.56.  What is Ka for the acid?

I don't understand this problem.

   HA        = A-    +    H+
.0234-x      0+x       .00275 +x


pH=2.56 [H+]=10^-2.56=.00275

[Demotivator]

The statement tells you that after dissociation of HA, [H+] is .00275,  not .00275+x.
So, guess what A- and HA would have to be by stoichiometry.

[AWK]

The statement tells you that after dissociation of HA, [H+] is .00275,  not .00275+x.

This problem needs more precise calculation
Ka=[A-][H+]/(c-[H+])
Neglecting dissociation of water: [A-] = [H+]
hence Ka=[H+]^2/(0.0243-[H+])

[Demotivator]

This problem needs more precise calculation
Ka=[A-][H+]/(c-[H+])
Neglecting dissociation of water: [A-] = [H+]
hence Ka=[H+]^2/(0.0243-[H+])

That's exactly what I was implying!