Untitled - Kelly Walsh High School

Buffers and Other Equilibria 241
16-5 Complex Ion Equilibria (K f)
We can treat other types of equilibria in much the same way as the ones previously
discussed. For example, there is an equilibrium constant associated with the formation
of complex ions. This equilibrium constant is the formation constant, K f.
Zn(H 2O) 4 2 reacts with ammonia to form the Zn(NH3) 4 2 complex ion according
to the following equation:
Zn(H 2O) 4 2 (aq) 4 NH3(aq) K Zn(NH 3) 4 2 (aq) 4H2O(l)
The K f of Zn(NH 3) 4 2 (aq) is 7.8 10 8 , indicating that the equilibrium lies to the right.
16-6 Utterly Confused About
pH and Titrations
Let’s examine a typical titration problem. A homework or exam problem may
utilize only one of the steps we will see in this problem, or you may need to
work all these steps. The goal of each step is to determine the pH of the solution.
You should be careful not to lose sight of this goal.
In many cases, you may know the initial concentration of the weak acid, but may
be interested in the pH changes during the titration. In order to do this you can
divide the titration curve into four distinctive areas in which the pH is calculated.
This first section is the initial pH. This is the point in the titration preceding the
addition of any reactant. You must ignore the reactant to be added in this region.
If a strong acid or a strong base is present, there are no significant equilibria. The
strong acid gives the hydrogen ion concentration directly, while a strong base gives
the hydroxide ion concentration directly. If a weak acid is present, this is a
generic K a problem, and you can find the hydrogen ion concentration from this. If
a weak base is present, this is a generic K b problem, and you can find the hydroxide
ion concentration from this. The hydrogen ion concentration directly gives you
the pH, while the hydroxide ion concentration will give you the pH indirectly.
In all other portions of the titration curve, you must consider both the substance
already in the container and the amount added. Calculations in these regions
will begin as a limiting reagent problem. The results of the limiting reagent calculation
will tell you how to finish the problem.
To do a limiting reagent problem you need to know how many moles of reactant
are in the original container and how many moles of the other reactant