|Similar calculations are used for weak bases.
|eg. NH3(g) + H2O(l)
<---------> NH4+(aq) + OH-(aq)
|This tells you that the dissociation constant for any of
bases can be obtained from the value of Kw divided by the
value of Ka. (Ka can also be obtained by
Kw by Kb).
|eg. NH3 Kw = 1.0 x 10-14
for NH4+ = 5.6 x 10-10
x 10-14 = 1.8 x 10-5
|What is the [OH-] of a 0.10 mol/L solution of
|When NaCN dissolves in water it dissolves completely
is a sodium salt.
NaCN -------> Na+ +
0.1 M 0.1 M 0.1 M
|However the CN- ions that are produced then
water and set up an equilibrium.
+ H2O <--------> HCN(aq) + OH-
0.1 M -x x x
|When we set up the equilibrium equation we must set it us
as a Kb
equation. Ka would be used if HCN was reacting with water.
time the salt of HCN, the CN- is reacting and it is acting
a base. Therefore we must use the Kb value.
|2.04 x 10-5 = (x)(x)
0.1 - x
x2 +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
the [OH-] = 1.42 x 10-3 mol/L.
Calculate the pH of a 0.10 mol/L NaCH3COO solution?