No, no ICE chart needed. Just remember this.
Ka*Kb = Kw = 1E-14
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?
2 answers
Thank you Dr. Bob222