Weak Bases 
Similar calculations are used for weak bases.

eg. NH_{3}(g) + H_{2}O(l)
<> NH_{4}^{+}(aq) + OH^{}(aq)

OR

This tells you that the dissociation constant for any of
the conjugate
bases can be obtained from the value of K_{w }divided by the
appropriate
value of K_{a}. (K_{a }can also be obtained by
dividing
K_{w }by K_{b}).

eg. NH_{3} K_{w }= 1.0 x 10^{14
}K_{a
}for NH_{4}^{+ }= 5.6 x 10^{10}
K_{b }=
1.0
x 10^{14 } = 1.8 x 10^{5}

FollowUp Problem 
What is the [OH^{}] of a 0.10 mol/L solution of
NaCN?

When NaCN dissolves in water it dissolves completely
because it
is a sodium salt.

NaCN > Na^{+ }+
CN^{}
0.1 M 0.1 M 0.1 M 
However the CN^{} ions that are produced then
react with
water and set up an equilibrium.

CN^{}(aq)
+ H_{2}O <> HCN(aq) + OH^{}
0.1 M x x x 
When we set up the equilibrium equation we must set it us
as a K_{b}
equation. K_{a} would be used if HCN was reacting with water.
This
time the salt of HCN, the CN^{} is reacting and it is acting
as
a base. Therefore we must use the K_{b} value.


2.04 x 10^{5} = (x)(x)
0.1  x x^{2 }+2.04 x 10^{5 }x 2.04 x 10^{6} = 0 Apply the quadratic formula and solve for 'x'. x = 1.42 x 10^{3}
mol/L Therefore
the [OH^{}] = 1.42 x 10^{3} mol/L.

FollowUp Problems
Calculate the pH of a 0.10 mol/L NaCH_{3}COO solution? 