PART A Stock Solution 10.0 mL of the 0.030 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) = 0.010 M KI = 0.010 M I- EXP 10-1 (of 8)

PART B Standard Solutions Tube # Stock I- Solution (mL) 0.20 M KNO3 (mL) 6 5.00 5.00 (Stock I- Solution: 0.010 M I-) MCVC = MD _______ VD = (0.010 M)(5.00 mL) _______________________ (10.00 mL) = 0.0050 M I- EXP 10-2 (of 8)

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

PART B Standard Solutions y = 215x + 0.028 A-456.0 = mx + b m(slope): 215 b(y-intercept): 0.028 A = (215 M-1)[I-] + 0.028 EXP 10-4 (of 8)

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

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

POST LAB QUESTION – FIND THE Ksp OF LEAD (II) IODIDE y = 215x + 0.028 A = (215 M-1)[I-] + 0.028 A - 0.028 = (215 M-1)[I-] A - 0.028 = [I-] ______________ 215 M-1 0.243 - 0.028 ___________________ 215 M-1 = 0.215 _________ 215 M-1 = 0.00100 M I- = [I-]eq EXP 10-7 (of 8)

POST LAB QUESTION – FIND THE Ksp OF LEAD (II) IODIDE PbI2 (s) ⇆ Pb2+ (aq) + 2I- (aq) Initial M’s Change in M’s Equilibrium M’s 0.0050 0.0050 - x - 2x 0.0050 - x 0.0050 - 2x 0.00100 0.00100 = 0.0050 – 2x –0.0040 = – 2x 0.0020 = x [Pb2+]eq = 0.0050 M – 0.0020M = 0.0030 M Ksp = [Pb2+][I-]2 = (0.0030 M)(0.00100)2 = 3.0 x 10-9 EXP 10-8 (of 8)