Textbook 6D #5

[Kailyn Missaghi]

Hi everyone, I am wondering for these types of questions how I go about solving this.

For a, first I made the equation NH3 + H2O --> NH4+ + OH-

Then I did the ice box so that it is 0.057-x on the reactants side and x for both products.

When doing Kb, I did, Kb= x^2/ 0.057-x, where do I go from there to get PH, POH, and % protonated?

[Austin Cortes 1E]

Hello, once you solve for x, you can determine the H+ (or OH-) concentration. By taking the -log of this, you can get the pH. In regards to the percent ionization, you would divide the final ion concentration by the initial concentration of the acid/base and then multiply by 100%.

[Beverly Lauring 1L]

I think these equations will be helpful to solve this question.

pOH=-log[OH-]
pH=14-pOH
% protonation = (equilibirum concentration of NH4+ / initial concentration of NH3) x 100%