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

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.

To calculate the equilibrium constant (K) for the reaction, you need to determine the concentrations of the reactants (CH3NH2 and H2O) and the products (CH3NH3+ and OH-) at equilibrium. Since you're given the initial concentration of CH3NH2 and the concentration of CH3NH3+ at equilibrium, you can use the stoichiometry of the balanced equation to find the concentration of OH- at equilibrium.

Let's denote the initial concentration of CH3NH2 as [CH3NH2]0, the equilibrium concentration of CH3NH2 as [CH3NH2]eq, and the concentration of OH- at equilibrium as [OH-]eq. The concentration of H2O remains essentially constant throughout the reaction and doesn't need to be considered in this case.

From the balanced equation:
CH3NH2 + H2O → CH3NH3+ + OH-

The stoichiometry tells us that the ratio of the concentration of CH3NH2 to OH- is 1:1. So, at equilibrium, [CH3NH2]eq = [OH-]eq.

Given that [CH3NH2]0 = 7.05×10-4 M and [CH3NH3+]eq = 0.000379 M, we can calculate the equilibrium concentration of CH3NH2:

[CH3NH2]eq = [OH-]eq = [CH3NH2]0 - [CH3NH3+]eq
= 7.05×10-4 M - 0.000379 M
= 6.67×10-4 M

Now, we can calculate the equilibrium constant (K) using the concentrations of the reactants and products at equilibrium:

K = ([CH3NH3+][OH-]) / ([CH3NH2][H2O])

Since the concentration of H2O doesn't change significantly, it can be assumed to be constant and omitted from the equation. Thus, the equation can be simplified to:

K = [CH3NH3+] / [CH3NH2]

Plugging in the values we have:

K = 0.000379 M / 6.67×10-4 M
= 0.568

Therefore, the equilibrium constant (K) for the reaction is approximately 0.568.