1. PART A Stock Solution
10.0 mL of the M KI in 0.20 M KNO3
and
20.0 mL of 0.20 M KNO3
MCVC = MD
_______ VD
= (0.030 M)(10.0 mL)
________________________
(30.00 mL)
= M KI
= M I-
EXP 10-1 (of 8)

2. PART B Standard Solutions
Tube # Stock I- Solution (mL) M KNO3 (mL)
(Stock I- Solution: M I-)
MCVC = MD
_______ VD
= (0.010 M)(5.00 mL)
_______________________
(10.00 mL)
= M I-
EXP 10-2 (of 8)

3. PART B Standard Solutions
4H I NO2- → I NO + 2H2O I- I2 I- I- I2 I-
EXP 10-3 (of 8)

4. PART B Standard Solutions
y = 215x
A = mx + b
m(slope): 215
b(y-intercept):
A = (215 M-1)[I-]
EXP 10-4 (of 8)

5. PART C Equilibrium Solutions
EXP 10-5 (of 8)

6. POST LAB QUESTION – FIND THE Ksp OF LEAD (II) IODIDE
Pb2+ (aq) I- (aq)
0.0050
0.0050
Pb2+
Pb2+
4H I NO2- → I NO + 2H2O I2 I- I-
PbI2
Read the absorbance of I2 from spectrophotometer at 456 nm :
0.243
EXP 10-6 (of 8)

7. POST LAB QUESTION – FIND THE Ksp OF LEAD (II) IODIDE
y = 215x
A = (215 M-1)[I-]
A = (215 M-1)[I-]
A = [I-]
______________
215 M-1
___________________
215 M-1 =
_________
215 M-1
= M I-
= [I-]eq
EXP 10-7 (of 8)

8. POST LAB QUESTION – FIND THE Ksp OF LEAD (II) IODIDE
PbI2 (s) ⇆ Pb2+ (aq) I- (aq)
Initial M’s
Change in M’s
Equilibrium M’s
0.0050
0.0050
- x
- 2x x x
Ksp = [Pb2+][I-]2
= ( M)( )2
= – 2x
= x 10-9
– = – 2x
= x
[Pb2+]eq = M – M
= M
EXP 10-8 (of 8)