The quantities you provide must meet the following criteria measured and calcula
ID: 503040 • Letter: T
Question
The quantities you provide must meet the following criteria measured and calculated guantifes 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 discussion PbCl2 (in water) mL Volume of PbCl2 (for one trial) mL Volume of AgNO3 (for one trial) mL (filtering rinsing three trials) Total volume of PbCl2 mL (rinsing three trials) Total volume of AgNO3 PbCI2 (in 0.0100 M Pb(NO3)2) mL. Volume of PbCl2 (for one trial) mL. Volume of AgNO3 (for one trial) mL (filtering rinsing three trials) Total volume of PbCl2 mL (rinsing three trials) Total volume of AgNO3 PbCl2 (in 0.100 M NaCl) mL. Volume of PbCl2 (for one trial Volume of AgNO3 (for one trial) mL mL (filtering rinsing three trials) Total volume of PbCl2 mL (rinsing three trials) Total volume of AgNO3Explanation / Answer
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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