Asked by Sam
Ammonia, NH3 is freely soluble in water, and has the value Kb=1.8x10^-5. If the pH of an ammonia solution is 11.4272, then what original concentration of NH3 (molarity) was used to make up the solution?
Answers
Answered by
DrBob222
NH3 + H2O ==> NH4^+ + OH^-
Kb = (NH4^+)(OH^-)/(NH3)
(x)(x)/(NH3)-x = 1.8 x 10^-5
pH = 11.4272 (where did all these places come from---we only know Kb to 2 places.)
pH = 11.43 pOH = 14-11.43 = 2.57
2.57 = -log(OH^-) and (OH^-) = 2.69 x 10^-3
Substitute into Kb expression and solve for (NH3). You should get something close to 0.4 M but you need to confirm all of the above.
Kb = (NH4^+)(OH^-)/(NH3)
(x)(x)/(NH3)-x = 1.8 x 10^-5
pH = 11.4272 (where did all these places come from---we only know Kb to 2 places.)
pH = 11.43 pOH = 14-11.43 = 2.57
2.57 = -log(OH^-) and (OH^-) = 2.69 x 10^-3
Substitute into Kb expression and solve for (NH3). You should get something close to 0.4 M but you need to confirm all of the above.
Answered by
Sam
Thank you! I tried all the above and got 0.40 M
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.