Let's look at some examples involving equilibrium concentrations of a weak base solution. What is the hydroxide concentration in a 0.150 molar NH3 solution? KB equals 1.8 times 10 the minus 5. NH3 is a base and we know it's a weak base. If we weren't sure, we'd see the Kb value and that indicates that we're dealing with a weak base solution. We're given the reaction of a aqueous ammonia plus water is an equilibrium within NH4 plus and Minus. Since we're dealing with a weak base, we're going to have to set up an ICE table, where we have I for initial, C for change and E For equilibrium. We're given the initial concentration of NH3 is 0.150, water is a pure liquid, so we don't have to worry about its concentration. Initially we have no NH4+, no Present. For the change row we have to look back at the stoichiometry of the balance chemical equation, so we have -x for the NH3 Plus x for NH4 plus, and plus x for Minus. For my equilibrium row, I simply sum the initial and change rows. So now I have my equilibrium concentrations with respect to x. Now I can set up the Kb expression. Kb, equal to the NH4 plus concentration times the Minus concentration divided by NH3 Concentration. The water is omitted from our Kb expression because it's a pure liquid. I now plug in the values I know, which are the Kb value is 1.8 x 10 to the -5. And if that's not given, that's a value you can generally look up. I have x for my NH4+ concentration, x for my Concentration and 0.150- X for the NH3 concentration. Now I'm going to make the assumption that x is much, much less than 0.150. This will allow me to simplify the expression to 1.8 x 10 to the -5 equals x squared over 0.150. Now I can solve for x, and x=1.6 times 10 to the minus 3. This will my hydroxide concentration. I also see that the value of x is only about 1% of the value of 0.150. So my assumption was valid. And I can find my hydroxide concentration just by using the value of X. It will also be the concentration of NH4+ since those two concentrations are equal.
