Chemistry

Given that Ka for HIO is 3.2 × 10-11 at 25 °C, what is the value of Kb for IO– at 25 °C? Then, Given that Kb for C6H5NH2 is 1.7 × 10-9 at 25 °C, what is the value of Ka for C6H5NH3 at 25 °C?

I don't know where to start? Do I need an ICE table?

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  1. No, no ICE chart needed. Just remember this.
    Ka*Kb = Kw = 1E-14

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  2. Thank you Dr. Bob222

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