Use ionization constants to calculate the.........

Use ionization constants to calculate the equilibrium constant for this overall reaction:
Ionizaion constants Ka of Acid:
HF: 6.3×10^-4, HClO: 4.0×10^-8

HF(aq) + CIO^-(aq) <-----> F^-(aq) + HClO(aq)

Let the equilibrium constant for the reaction be called Kr. Let the Ka's be called Ka(HF) and Ka(HClO).

Kr = [F-][HClO]/[HF][ClO-]

The data you have to work with:

Ka(HF) = [H+][F-]/[HF] = 6.3 x 10^-4

Ka(HClO) = [H+][ClO-]/[HClO] = 4.0 x 10^-8

The first thing to notice about Kr is that you have to have [F-] in the numerator and [HF] in the denominator. To get that you have to use Ka(HF) as written.

The second thing to notice is that you have to have [HClO] in the numerator and [ClO-] in the denominator. To get that you have to divide Ka(HF) by Ka(HClO). To divide by a fraction, invert the fraction and multiply. After that, cancel [H+] in numerator and denominator to get:

Kr = Ka(HF)/Ka(HOCl) = [H+][F-][HClO]/[HF][H+][ClO-] = (6.3x10^-4)/(4.0x10^-8) = 1.6 x 10^4

Yes, that's right: 16,000