how do you find pH of an acid/base

My memory of chemistry class fails me, and I know this isn't a chem forum, but I was hoping you can help me with the math part of a certain problem.

How do you figure out the pH of a base, given only the molarity and the Kb value. ie: .10M NH4Cl

I appreciate any help.

Its -log₁₀ of the hydrogen ion concentration which is about 10^-7 for pure water (hence pH 7 is neutral).

You need to assume 100% dissociation of your NH₄Cl...

hmm...i think i now remember. since my problem was given to me in a solution, i would need to figure out the molarity in the solution, not the pure substance, and then take the negative log of that. i was confused by all the Ka and Kb values. thanks for the help.

In that situation:

K_b = [NH₄⁺][OH^-] / [NH₃]

Aqueous ammonia is only slightly ionized so it's concentration is effectively equal to the original concentration of the solution, i.e. [NH₃] = 0.1M. The base ionization constant for ammonia is 1.6 x 10^-5, so:

1.6 x 10^-5 = [NH₄⁺][OH^-] / 0.1

The chemical equation for the equilibrium is:
NH₃ + H₂O <----> NH₄⁺ + OH^-
which indicates the concentrations of ammonium and hydroxide ion are equal, i.e. [NH₄⁺] = [OH]^-, so:

1.6 x 10^-5 = [OH^-]² / 0.1
[OH^-]² = 1.6 x 10^-6
[OH^-] = 1.3 x 10^-3.

The ionic product of water requires that:
[H₃O⁺][OH^-] = 1 x 10^-14
Substitute from above:
[H₃O⁺] x 1.3 x 10^-3 = 1 x 10^-14
giving:
[H₃O⁺] = 7.7 x 10^-12M

Now just stick that into the formula:
pH = -log₁₀[H₃O⁺] = -log₁₀(7.7 x 10^-12)
= 11.1 (1.d.p)

Ta-da.

sweet...thanks