The estimated [AgNO3] provided is 0.050 M. Chem 1132 Lab Ksp of Lead(lI) Chlorid
ID: 548394 • Letter: T
Question
The estimated [AgNO3] provided is 0.050 M.
Chem 1132 Lab Ksp of Lead(lI) Chloride Pre-lab answers due in lecture on Wednesday preceding the laboratory Name The quantities you provide must meet the following criteria (a) measured and calculated quantifies have four significant figures, (b) only 5.000-, 10.00-, and 25.00-mL volumetric pipets available (c) use a minimum volume of the saturated lead(I) chloride. Ksp PbCl2 (s) theoretical Estimated [AgNO3] provided in pre-lab material PbCl2 (in water) Volume of PbCl2 (for one trial) Volume of AgNO3 (for one trial) Total volume of PbCl2 Total volume of AgNO3 mL mL mL (filtering+ rinsing + three trials) mL (rinsing+ three trials) PbCl2 (in 0.0100 M Pb(NO3)h) mL Volume of PbCl2 (for one trial) Volume of AgNOs (for one trial) Total volume of PbClh Total volume of AgNO3 mL mL (filtering+ rinsing+ three trials) mL (rinsing + three trials) PbCl2 (in 0.100 M NaCI) mL Volume of PbCl2 (for one trial) Volume of AgNOs (for one trial) Total volume of PbCl2 Total volume of AgNOs mL mL (filtering + rinsing+ three trials) mL (rinsing+three trials)Explanation / Answer
SOLUTION:
Ksp for PbCl2= 1.7 x 10-5 , Std AgNO3 Solution = 0.1 M
1. First Case : PbCl2 in water (molar mass of PbCl2= 278.1 gm)
PbCl2 = Pb2+ + 2Cl-
Ksp = [Pb2+] [Cl-]2 = [x]. [x]2 = 1.7 x 10-5 = x3
x = 0.005 M PbCl2
Solubility of PbCl2 = 0.005 * 278.1 = 1.39 gm/L
Volume of PbCl2 taken by 25ml pipette = 10ml (Suppose)........First trial
0.005 x 25 = 0.1 VAgNO3
Volume of 0.1 M AgNO3 required will be = 1.25 ml
Suppose rinsing of glassware takes 5ml solution and filteration requires 5 ml solution.
Total Volume of PbCl2 = Filtering + 3 trial + 3 rinsing = 5 +15 +75 = 95 ml (approx.)
Total Volume of AgNO3 = 3 times rinsing + 3 trial = 15 + 3.75 = 19 ml (approx)
2. Second Case : PbCl2 in 0.01M Pb(NO3)2
PbCl2 = Pb2+ + 2Cl-
0.01M Pb(NO3)2 gives 0.01 M Pb2+ ions in solution.
Ksp = [Pb2+] [Cl-]2 = 1.7 x 10-5 = 0.01 [Cl-]2
[Cl-]2 = 1.7 x 10-3 M
[Cl-] = 0.41 x 10-2
Volume of PbCl2 taken by 25ml pipette = 10ml (Suppose)........First trial
0.41 x 10-2 x 25 = 0.1 V
Volume of 0.1 M AgNO3 required will be = 1.025 ml
Total Volume of PbCl2 = Filtering + 3 trial + 3 rinsing = 5 +15 +75 = 95 ml (approx.)
Total Volume of AgNO3 = 3 times rinsing + 3 trial = 15 + 3.075 = 19 ml (approx)
2. Third Case : PbCl2 in 0.1M NaCl
PbCl2 = Pb2+ + 2Cl-
0.1M NaCl gives 0.1 M Cl- ions in solution.
Ksp = [Pb2+] [Cl-]2 = 1.7 x 10-5 = [Pb2+] 0.1*0.1 = [Pb2+] 0.01
[Pb2+] = 1.7 x 10-3 M
[Cl-] = 3.4 x 10-3 M
Volume of PbCl2 taken by 25ml pipette = 10ml (Suppose)........First trial
3.4 x 10-3 x 25 = 0.1 V
Volume of 0.1 M AgNO3 required will be = 0.85 ml
Total Volume of PbCl2 = Filtering + 3 trial + 3 rinsing = 5 +15 +75 = 95 ml (approx.)
Total Volume of AgNO3 = 3 times rinsing + 3 trial = 15 + 3 = 18 ml (approx)
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