Untitled - Kelly Walsh High School

Acids and Bases 229
We do not have a reaction given, therefore we must write one. The “b” subscript
in the K b tells us the equation must look like this:
NH 3(aq) H 2O(l) K NH 4 (aq) OH (aq)
We need to write the equilibrium constant expression for this reaction:
We can now create an ICE table:
K b
NH 3 OH NH 4
Initial 0.500 0 0
Change x x x
Equilibrium 0.500 x x x
We can enter the values from the equilibrium line into the equilibrium constant
expression:
[NH4 ] [OH ] [x] [x]
Kb
[NH3 ] [0.500 x]
There is one additional piece of information in the problem, and that is the pH.
From the pH we can determine the hydrogen ion concentration:
pH 11.48
[H ] 1011.48 [H ] 3.3 10 12 M
From the hydrogen ion concentration and K w we can determine the hydroxide
ion concentration in the solution.
K w [H ][OH ] 1.00 10 14
[OH ] 3.0 103 M
[H 1.00 1014
] 3.3 1012 From our table, we know that the hydroxide ion concentration is x. Therefore,
we can substitute 3.0 10 3 into the equilibrium expression for x and enter the
values into a calculator:
[NH 4 ] [OH ]
K w
[NH 4 ] [OH ]
[NH 3 ]
Kb 1.8 105 [3.0 103 ] [3.0 103 ]
[0.500 3.0 103 [x] [x]
[NH3 ] [0.500 x]
]