solve for x to determine the [OH-] for 0.25 M methylamine, CH3NH2

Question

[OH-] == x

[CH3NH2] == 0.25 - x

[CH3NH3+] == x

Kb = 4.00 x 10^-4

x == 4.00x10^4 * (0.25-x)/x

Answer 1

OK, starting from K_B = x²/(.25-x) = 4 x 10^-4 (You have the K expression correct), you should solve the quadratic equation in x. This is why we usually assume the concentration of the weak acid or base is >> x, the ionization. (This leads to x = sqrt(K_B/.25) or 0.01 M which is 4% ionization)

Multiplying by the denominator in the fraction and rearranging:

x² + (4x10^-4)x - 10^-4 = 0

Now use quadratic formula: (-.0004 ±sqrt( (.0004)² +.0004)) / 2 = .0098 M

Note that the approximation that .25 >> x is still reasonable with K = 4x10^-4 (about 2% error). There are few weak acids or bases that solving the full quadratic is necessary.

Take care