Scenario:A student mixed 5.0 mL of 0.015 M Pb(NO 3 ) 2 with 3.0 mL of 0.030 M KI in 2.0 mL
of 0.200 M KNO 3. The test tube was shaken for 15 minutes and the solid precipitate of lead (II)
iodide was allowed to settle. The tube was centrifuged to remove any excess lead (II) iodide
from the supernatant. To the supernatant, she added KNO 2 to oxidize the I–to I 2. The super-
natant was then analyzed for I–by using a spectrophotometer. She then made known dilutions
of a 0.10 M potassium iodide solution in acidified KNO 2 to create a calibration curve to be able
to determine the I–in the supernatant.

Analysis:

1. How many moles of Pb2+were initially present?
   .'
   '

'

'

mL sol n .()

mL sol n

L sol n

L sol n

mol Pb NO

1 1000

1

1

(^50) 0 015 32

= 7.5 × 10 –5mol Pb(NO 3 ) 2 yielding 7.5 × 10 –5mol Pb2+

2. How many moles of I–were initially present?
   .'
   '

'

'

mL sol n.

mL sol n

L sol n

L sol n

mol KI

1 1000

1

1

(^30) 0 030

= 9.0 × 10 –5mol KI yielding 9.0 × 10 –5mol I–

3. From known dilutions of KI in acidified KNO 2 , a calibration curve was created. From the
   calibration curve, it was determined that the concentration of I–at equilibrium in the su-
   pernatant was 1.4 x 10–3mole/liter. How many moles of I–were present in 10. mL of the
   final solution?
   .^3. .'. 5
   L

moles L mol I in sol n at equil

1

14 10 0 010

# ##=14 10

4. How many moles of I–precipitated?
   moles I–originally present −moles of I–in solution
   = 9.0 × 10 –5− 1.4× 10 –5= 7.6 × 10 –5moles I–ppt.


5. How many moles of Pb2+precipitated?
   There are twice as many moles of I–as there are Pb2+, therefore,
   . + 5 +
   .

mol I..

mol I

mol Pb moles Pb

1

76 10

200

(^100) 38 10
(^52)

##= 2

--

6. How many moles of Pb2+were in the solution?
   total moles of Pb2+−moles of Pb2+in ppt
   = 7.5 × 10 –5−3.8 × 10 –5= 3.7 × 10 –5moles Pb2+in sol’n

Part III: AP Chemistry Laboratory Experiments