Cliffs AP Chemistry, 3rd Edition

and results in two separate equilibrium expressions:

[]

K [][]

HA

HO HA

a1=^32

+-

[]

K [][]

HA

HO A

a2^3

2

= -

+-

Figure 2

Scenario: A student measured out approximately 10 mL of 6.00 M NaOH and diluted the base
to approximately 600 mL. The student then performed an acid-base titration (see Figure 3) and
determined that 48.7 mL of NaOH solution were needed to neutralize 50.0 mL of a 0.100 M
HCl solution.

Once the student had determined the exact concentration of the base, the student then pro-
ceeded to determine the equivalent mass of an unknown acid. To do this, the student measured
out 0.500 grams of an unknown solid acid and titrated it with the standardized base, recording
pH with a calibrated pH meter as the base was added. The student added 43.2 mL of the base
but went too far past the end point and needed to back-titrate with 5.2 mL of the 0.100 M HCl
to exactly reach the end point.

Analysis:

1. Calculate the molarity of the NaOH solution from Part I.

.

.

mL mL..

mol HCl

mol HCl

mol NaOH

L M NaOH

50 0^1

1000

0 100

1

1

###0 0487 =0 103

2. The graph of the titration from this experiment is presented here. Determine the Kaor
   Ka’s.

pH

Volume of NaOH added (mL)

0 20 40 60 80

pH when half of 1st H+

is neutralized

pH when 2nd H+ is neutralized

pH when half of 2nd H+

is neutralized

pH when 1st H+ is neutralized

Part III: AP Chemistry Laboratory Experiments