5 Steps to a 5 AP Chemistry

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Alternate solution:

Titration Equilibria

An acid–base titrationis a laboratory procedure commonly used to determine the concentra-

tion of an unknown solution. A base solution of known concentration is added to an acid

solution of unknown concentration (or vice versa) until an acid–base indicatorvisually

signals that the end pointof the titration has been reached. The equivalence pointis the

point at which a stoichiometric amount of the base has been added to the acid. Both chemists

and chemistry students hope that the equivalence point and the end point are close together.

If the acid being titrated is a weak acid, then there are equilibria which will be established

and accounted for in the calculations. Typically, a plot of pH of the weak acid solution being

titrated versus the volume of the strong base added (the titrant) starts at a low pH and grad-

ually rises until close to the equivalence point, where the curve rises dramatically. After the

equivalence point region, the curve returns to a gradual increase. This is shown in Figure 15.3.

In many cases, one may know the initial concentration of the weak acid, but may be

interested in the pH changes during the titration. To study the changes one can divide the

titration curve into four distinctive areas in which the pH is calculated.

1. Calculating the initial pH of the weak acid solution is accomplished by treating it as a

simple weak acid solution of known concentration and Ka.

2. As base is added, a mixture of weak acid and conjugate base is formed. This is a buffer

solution and can be treated as one in the calculations. Determine the moles of acid con-

sumed from the moles of titrant added—that will be the moles of conjugate base

formed. Then calculate the molar concentration of weak acid and conjugate base, taking

into consideration the volume of titrant added. Finally, apply your buffer equations.

3. At the equivalence point, all the weak acid has been converted to its conjugate base. The

conjugate base will react with water, so treat it as a weak base solution and calculate the

[OH−] using Kb. Finally, calculate the pH of the solution.

4. After the equivalence point, you have primarily the excess strong base that will deter-

mine the pH.

pOH

NH

NH ] 3

=− × +

=+

−

+

log. log

[]

[

.log

.

181 10

4 742

300

2

(^54)

..

.

00

==4 918 pH 9.082

181 10

300

121 10

14 00

5

5

.

.

.

.

×=

=×

=

=

−

−

x

x

2.00

pOH 4.918

pH 0 0 4 918 9 082-.=.

K

xx

b x

4

+

3

NH OH

NH ]

)( )

)

==

+

−

=×

[][]−

[

(.

(.

.

300

200

181 10−−^5

224  Step 4. Review the Knowledge You Need to Score High