Asked by harry
Calculate the equilibrium constant for the weak base CH3NH2, if a solution of the base with an initial concentration of 7.05×10-4 M has a [CH3NH3+] of 0.000379 M (make an exact calculation assuming that initial concentration is not equal to the equilibrium concentration).
CH3NH2 + H2O = CH3NH3+ + OH-
i don't know how to do this problem help is appreciated
CH3NH2 + H2O = CH3NH3+ + OH-
i don't know how to do this problem help is appreciated
Answers
Answered by
DrBob222
I'm not exactly sure about this problem because I don't understand the portion in the parentheses. My assumption is that the system has not reached equilibrium (after-all that's what the problem states) so the concns are as listed.
So I would write the Kb expression.
Kb = (CH3NH3^+)(OH^^-)/(CH3NH2)
and plug in the values given.
If CH3NH3^+ is 0.000379, that must be the concn of OH^-, too, and CH3NH2 must be what we started with minus the amount of CH3NH3^+ formed or 0.000705-0.000379. I worked through the problem and obtained 4.41 x 10^-4 for Kb. My OLD quant book (copyright 1992) gives a value of 4.4 x 10^-4. Fairly close. Check my thinking.
So I would write the Kb expression.
Kb = (CH3NH3^+)(OH^^-)/(CH3NH2)
and plug in the values given.
If CH3NH3^+ is 0.000379, that must be the concn of OH^-, too, and CH3NH2 must be what we started with minus the amount of CH3NH3^+ formed or 0.000705-0.000379. I worked through the problem and obtained 4.41 x 10^-4 for Kb. My OLD quant book (copyright 1992) gives a value of 4.4 x 10^-4. Fairly close. Check my thinking.
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.