A 0.057M HIO3 solution is 10.0% ionized.
What is the dissociation constant (ka) for HIO3? And what is the pOH of the solution?

The dissociation constant is given by:

K_a = \frac{[H^+ (aq)][IO_3\ ^- (aq)]}{[HIO_3 aq]}

Work out the number of moles of H+ and IO3- formed, from the given numbers, and the HIO3 left.

From that, you can find each concentration, and thus the constant.

Do you know how to find pH?

pH = -log\ [H^+]

Well, pH + pOH = 14, you can find pOH.

The answer that I got was: a) 6.3*10^-4
b) and the pOH is 11.18

While I got the constant, I got the pH as 11.75.

[H^+] = 5.7 x 10^-3

So, pH = - log [5.7 x 10^-3] = 2.244125144

pOH = 14 - 2.24 = 11.75597... = 11.76