Untitled - Kelly Walsh High School

230 CHEMISTRY FOR THE UTTERLY CONFUSED
Don’t Forget!
In our final example, we will do a problem with a twist.
Determine the pH of a 0.200 M strontium acetate solution.
Strontium acetate is neither a weak acid nor a weak base—it is a salt. As a soluble
salt, it is a strong electrolyte and it will dissociate as follows:
Sr(C 2H 3O 2) 2(aq) l Sr 2 (aq) 2 C 2H 3O 2 (aq)
The resulting solution has 0.200 M Sr 2 (we do not need this value) and 2(0.200 M)
0.400 M C 2H 3O 2 .
We can ignore ions such as Sr 2 , which come from strong acids or strong bases
in this type of problem. Ions, such as C 2H 3O 2 , from a weak acid or a base, weak
acid in this case, will undergo hydrolysis, a reaction with water. The acetate ion
is the conjugate base of acetic acid (K a 1.74 10 5 ). Since acetate is a weak
base, this will be a K b problem, and OH will form. The equilibrium is:
C 2H 3O 2 (aq) H2O(l) K OH (aq) HC 2H 3O 2(aq)
You can ignore ions from a strong acid or a strong base. You must work with ions
from a weak acid or a weak base.
The equilibrium constant expression for this reaction is:
K b
We can now create an ICE table:
[OH ] [HC 2 H 3 O 2 ]
[C 2 H 3 O 2 ]
C 2H 3O 2 OH HC 2H 3O 2
Initial 0.400 0 0
Change x x x
Equilibrium 0.400 x x x
We can enter the values from the equilibrium line into the equilibrium constant
expression:
[OH [x] [x]
Kb
[0.400 x]
] [HC2H3O2 ]
[C2H3O2 ]