Asked by Nancy
A 4.50e-2 M solution of unknown monoprotic weak base has a pH of 10.200 what is the value of kb for the base
Answers
Answered by
DrBob222
BOH ==> B^+ + OH^-
Kb = (B^+)(OH^-)/(BOH)
pH = 10.2; therefore, pOH = 3.8
pOH = -log(OH^-). Solve for OH^- and substitute into Kb expression. For (BOH) substitute 0.045-(OH^-).
Post your work if you get stuck.
Kb = (B^+)(OH^-)/(BOH)
pH = 10.2; therefore, pOH = 3.8
pOH = -log(OH^-). Solve for OH^- and substitute into Kb expression. For (BOH) substitute 0.045-(OH^-).
Post your work if you get stuck.
Answered by
Anonymous
For pOH=-log(OH-) I solved to get OH=1.58e-4
then I did kb=(4.50e-2)(1.58e-4)/.045-1.58e-4 to get 1.59e-4 but that's not the right answer
then I did kb=(4.50e-2)(1.58e-4)/.045-1.58e-4 to get 1.59e-4 but that's not the right answer
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.