Question

I have a compound (mx)
weak acid (x- = Hx)
mx in 5M strong acid solution
solubility = 5x10(-3)
Ksp = 1x10(-20)
Find Ka...
I need to know where to start

Answer 1

You start here.
MX ==> M^+ + X^-
Ksp = (M^+)(X^-) = 1E-20
Then HX ==> H^+ + X^-
Ka = (H^+)(X^-)/HX) = ?

Note that X^- + H^+ ==>HX which is just the reverse of the ionization.
Note, also, that the addition of a strong acid (5M of a strong acid to boot), means that tha solubility of MX is increased. WHY? because adding H^+ causes the HX equilibrium to move to the right (X^- + H^+ => HX) and that makes MX solubility go to the right.
So S = solubility of MX, which means (M^+) = S and (X^-) + (HX) = S
So you don't know X, or HX, or Ka.