EXPERIMENTAL PROCEDURE Study this Sudy this sWear safety goggles when performing

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EXPERIMENTAL PROCEDURE Study this Sudy this sWear safety goggles when performing this experiment.) s section and the Pre-Laboratory Questions before coming to the laboratory n of Saturated Solutions of Cadmium Oxalate A. Preparation of Satu ned to prepare three saturated solutions of Cdc,o, Mark three small beakers 1,2, You . Clean, and properly rinse your buret (LABORATORY METHODS B). Fill the buret and 3. Clean, and with 0.100 NM h 0100 M cadmium nitrate, Cd(NO,), solution. To beaker l add 200. mL of wiu tr water and 25.00 mL of Cd(NO,), solution. To beakers 2 and 3 add 10.00-mL distilled w ns of Cd(NO,), solution from the buret. Clean your buret again, and fill it this th 0.100 M sodium oxalate, Na,C,O,, solution. Add 25.00 mL of this portions of with 0.100 NM time wi Na,C,O, so lution to beaker 1, and 10.00 mL each to beakers 2 and 3. Stir the mixtures ,solut for at least 10. minutes to allow the particle size to grow, and then permit the precipitates to settle. B. Determination of the Solubility Product of Cdc,o Using either gravity filtration or suction filtration (LABORATORY METHODS D), filter the mixture in beaker 1. Clean your buret, and rinse it with a few milliliters of the filtrate. Using your buret, measure 100. mL of filtrate into each of two Erlenmeyer flasks. You will need to fill your buret twice in order to measure the required 100. mL of filtrate for each of the two flasks. Add 20. mL of 3 M sulfuric acid, H,SO, solution to each flask. CAUTION: Dilute H,SO, is corrosive and causes burns on your skin or holes in your clothing. If there are any spils or spatters onto your skin or clothing rinse the affected area thoroughly with water. Clean, rinse, and fill your buret with approximately 0.01 M potassium permanganate, KMnO,, solution. Record in TABLE 19.1B the KMnO, molarity shown on the bottle eat one 100-mL portion of the filtrate to 55°C-60.°C, and itrate the hot solution with the KMnO, solution. The endpoint will be the point in the titration at which the pink color of KMno, first persists for about 30. seconds. Record your titration data in TABLE 19.1B. Repeat the titration with the second 100-mL portion of CaC,0, solution, and record your data in TABLE 19.1B.

Explanation / Answer

  in part A:
you added equal amounts of Cd+2 & C2O4)-2 when you added 25 ml of 0.1 molar solutions of each to the 200ml of water

when the titration finds the concentration of (C2O4)-2 ,... it also finds the concentratrion of Cd+2 left in solution after the precipitate is removed

first find moles of KMnO4:
0.0035 Litres KMnO4 @ 0.00997 mol/Litre = 3.4895 e-5 moles KMnO4

because the H2SO4 was added,... it tells us that the redox reaction is:
2 KMnO4 & 5 H2C2O4 & H2SO4 --> (& show you how if you care)
therfeor with that ratio:
3.4895 e-5 mol KMnO4 @ 2 mol (C2O4)-2 / 5 mol KMnO4 =1.396e-5 mol (C2O4)-2

now because this is the total amount of moles in a 100ml sample of solution ,... that original solution had a concentration before titration :
1.396e-5 mol (C2O4)-2 / 0.100 Litres = 1.396e-4 mol/L (C2O4)-2

CdC2O4 <=> Cd+2 & C2O4)-2

Ksp = [Cd] [C2O4]

Ksp = [1.396e-4] [1.396e-4]

Ksp = 1.948e-8

but unless you had more sig figs in the "3.5ml" of KMnO4,(ie if it was 3.50 ml instead), this became a 2 sigfig problem , & should be rounded off to

Ksp = 1.9e-8

which is cool because googling came up with 1.5e-8
http://www.google.com/search?sourceid=na...
======================================...

part B

you had the equilibrium:
CdC2O4 <=> Cd+2 & (C2O4)-2
by adding NH3, we say that "virtually all" of the moles of Cd+2
is converted into [Cd(NH3)4]+2 .... let's find the amount of
[Cd(NH3)4]+2 formed:

first find the moles of Cd+2 in the10ml of 0.1 M Cd(NO3)2:
0.010 Litres @ 0.100 mol/litre = 0.0010 moles Cd+2

0.0010 moles Cd+2 produced 0.0010 moles [Cd(NH3)4]+2

next , if 10 ml of Cd+2 solution & 10 ml 0f C2O4)-2 solution had 2 ml of NH3 solution mixed,... you have a total of 22 ml of solution ,.... find the molarity of [Cd(NH3)4]+2 :
0.0010 moles / 0.022 litres = 4.545e-2 Molar [Cd(NH3)4]+2
===================

now though "virtually all" of the moles of Cd+2
is converted into [Cd(NH3)4]+2 .... let's find the amount of Cd+2 that did not :

first find [C2O4-2]:
10 mls of 0.10 Molar Na2C2O4 has 0.0010 moles (C2O4)-2
it also was in that same 22 mls of mix:
0.0010 mol / 0.022 L = also 4.545e-2 Molar (C2O4)-2

CdC2O4 <=> Cd+2 & C2O4)-2

Ksp = [Cd] [C2O4]

1.948e-8 = [Cd] [4.545e-2 ]

[Cd+2] = 4.286e-7 Molar
======================

I am sure there might be a short cut in this next part,... but

we had 2 ml of 5 Molar NH3 diluted into 22 ml,...
making it 11 times weaker,... 0.4545 Molar NH3

a lot of NH3 was tied up in the 4.545e-2 Molar [Cd(NH3)4],..
@ 4NH3 in 1 [Cd(NH3)4]+2 ,... 0.1818 Molar NH3 is tied up in it,....

that leaves 0.4545 Molar NH3 - 0.1818 Molar NH3 tied up =
0.2727 Molar still in solution

if Cd+2 & 4 NH3 --> [Cd(NH3)4],.. then

K = [Cd(NH3)4] / [Cd] [NH3]^4

K = [4.545e-2] / [4.286e-7] [0.2727]^4

K = [4.545e-2] / [4.286e-7] (5.53e-3)

K = [4.545e-2] / 2.37e-9

K = 1.918e7

the 2.0 ml of NH3 reduced the sigfigs to 2 digits, but
the NH3 concentration of "5M" reduces the sigfigs to 1 digit.

so check the sig figs & choose either
2sigfigs @ K = 1.9e7
or 1 sigfig @ K = 2e7

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