Question

Calculate the initial molarity of KNH2 and the molarities of K+, NH3, OH-, and H3O+ in an aqueous solution that contains 0.75 g of KNH2 in 0.255 L of solution. You may ignore the reaction of NH3 with water.

I already found the initial molarity, but I am not sure how to find each species concentration with the information given. Please explain how to solve rather than give me only answers--they are helpful to compare but I want to learn how to do this. Thanks!
NH2^- + HOH ==> NH3 + OH^-
Set up ICE chart with NH3 = x and OH^- = x so NH2^- = your molarity -x
Solve for x.
Kb = (Kw/Ka) = (NH3)(OH^-)/(NH2^-)
That gives you NH3, OH, you have NH2^- and K^+ is same as NH2^-.
This could be a dumb question but how do you find Kb here in order to solve for x?

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