General

[quote=NonSonoFronz]@Neuro: Damn... Then, I don't know what the hell is wrong.
Because that's exactly how you find [OH-] given that the .1M was the initial concentration of CH3NH3Cl.

You could try finding the Ka given the Kb and solving it that way...[/quote]

found the answer, you have to find ka since the solution is acidic, then you get the answer, thanks anyways though

@Neuro: Damn... Then, I don't know what the hell is wrong.
Because that's exactly how you find [OH-] given that the .1M was the initial concentration of CH3NH3Cl.

You could try finding the Ka given the Kb and solving it that way...
Actually, definitely try doing that. Given the Ka value you'll probably come close to the answer you're looking for.

[quote=NonSonoFronz]CH3NH3Cl + H2O -> CH3NH4Cl+ + OH-

4.4x10^-4 = x^2/.10-x
x = [OH-]

Solve for x, calculate pOH, then subtract to find pH.[/quote]

answer is wrong if you do it that way, i get 2.17 for the pOH

CH3NH3Cl + H2O -> CH3NH4Cl+ + OH-

4.4x10^-4 = x^2/.10-x
x = [OH-]

Solve for x, calculate pOH, then subtract to find pH.