The Foundations of Chemistry

(Marcin) #1
EXAMPLE 18-9 Calculation of Kafrom pH
The pH of a 0.115 Msolution of chloroacetic acid, ClCH 2 COOH, is measured to be 1.92.
Calculate Kafor this weak monoprotic acid.
Plan
For simplicity, we represent ClCH 2 COOH as HA. We write the ionization equation and the
expression for Ka. Next we calculate [H 3 O] for the given pH and complete the reaction
summary. Finally, we substitute into the Kaexpression.
Solution
The ionization of this weak monoprotic acid and its ionization constant expression may be
represented as

HAH 2 O 34 H 3 OA and Ka

We calculate [H 3 O] from the definition of pH.

pHlog [H 3 O]

[H 3 O]^10 pH^10 1.920.012 M

We use the usual reaction summary as follows. At this point, we know the original[HA] and
the equilibrium[H 3 O]. From this information, we fill out the “change” line and then deduce
the other equilibrium values.

HA H 2 O^34 H 3 O  A
initial 0.115 M  0 M 0 M
change due to rxn 0.012 M 0.012 M 0.012 M
at equil 0.103 M 0.012 M 0.012 M

Now that all concentrations are known, Kacan be calculated.

Ka1.4 10 ^3

You should now work Exercise 40.

(0.012)(0.012)

0.103

[H 3 O][A]

[HA]

[H 3 O][A]

[HA]

764 CHAPTER 18: Ionic Equilibria I: Acids and Bases


Problem-Solving Tip:Filling in Reaction Summaries

In Examples 18-8 and 18-9 the value of an equilibrium concentration was used to deter-
mine the change in concentration. You should become proficient at using a variety of
data to determine values that are related via a chemical equation. Let’s review what we
did in Example 18-9. Only the equilibrium expression, initial concentrations, and the
equilibrium concentration of H 3 Owere known when we started the reaction summary.
The following steps show how we filled in the remaining values in the order indicated
by the numbered red arrows.
1.[H 3 O]equil0.012 M, so we record this value
2.[H 3 O]initial0, so change in [H 3 O] due to rxn must be 0.012 M
3.Formation of 0.012 MH 3 Oconsumes 0.012 MHA, so the change in [HA]
0.012 M
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