posted by Trixie on .
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. I know how to set up the ICE chart, but am unsure what to set my equation equal to.
The equation I am using for my ICE chart is set up as NH2^- + HOH --> NH3+OH^-....I am a little confused where I find the concentration of H3O+
(DrBob, I did see your explanation, and I see the general idea of how to proceed with this, but am stuck)
You solve that equation I gave you for x. That gives you (OH^-) and (NH3). Convert OH^- to pOH, then to pH, then to H3O^+.
Kb = (Kw/Ka) = (NH3)(OH^-)/(NH2^-)
(Kw/Kb) = (x)(x)/(your molarity -x)
One unknown only. Solve for x. x = (OH^-) = (NH3).
Right...I am confused on the solving for x part. Before when I have used this I have always been given ka or kb in order to solve for x. This time I'm not--so I don't know how to set up my equation to find x. Once I have x I know how to do everything else
Trixie--I apologize for not getting back sooner but I looked for Ka NH3 in all of my references and spent about three hours looking on the web but I couldn't find it. I can't believe I've been in the business this many years and I've not seen this before. There are a number of references that show Kb for NH2^- = Kw/Ka, and they calculate Ka as Kw/Kb which just makes Kb for NH2^- regenerate Kb for NH3. I don't think that is correct but I may be mistaken.
A key point to realize is that KNH_2 dissociates completely, and there is an excess of water. If you take that into account, there is no need for Kb or even an ICE chart. The concentration of the products will equal the concentration of the reactants. Hope that clears it up a bit, and hopefully this answer isn't too late to help, haha.