CHEM
posted by harry on .
Calculate the equilibrium constant for the weak base CH3NH2, if a solution of the base with an initial concentration of 7.05×104 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

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 (afterall 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 0.0007050.000379. 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.