What is the pH of .7 M (CH3NH3CL)? Kb=4.2*10^-7?

Answer 1

CH3NH3Cl completely dissociate into CH3NH3+ and Cl-. There is an equilibrium in the solution:
CH3NH3+ + H2O <==> H3O+ + CH3NH2, Ka
Ka for this reaction relates to Kb for:
CH3NH2 + H2O <==> CH3NH3+ + OH-, Kb
such that Ka*Kb = 10^-14.
Once you have Ka, pH can be calculated in the following way:
..CH3NH3+ + H2O <==> H3O+ + CH3NH2
Initial: 0.7._._._._._._._._.~0._._._.0.0
Final: 0.7 - X._._._._._._.X._._._._.X
when Ka is very small, X can be very small and hence 0.7 - X ~ 0.7, and thus
Ka = [H3O+]*[CH3NH2]/[CH3NH3+]
=X^2 /0.7
Therefore [H+] = X = sqrt(0.7*Ka)
and pH = -log([H+]) = -log(sqrt(0.7*Ka))
= -0.5*log(0.7*Ka) = 0.5*pKa - 0.5*log(0.7)
But I do not believe your Kb. Please check.