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**Unformatted text preview: **Lab #8: Determining a Solubility Product Constant
Prelab Questions:
1.
A) Calculate the mass of solid potassium iodate (KClO
of 0.200 M KClO 3 .
*50mL= 0.05 L & 0.200 M= 0.200 mol/L
Mass of KClO 3 = 0.05 L x 0.200 mol KClO 3 ,
1L x 3 , 214.00 g/mol) needed to prepare 50mL 214.00 g/mol
1 mol ≈ 2.1 g = 2.14 B) Calculate the mass of calcium nitrate tetrahydrate (Ca(N O 3 ¿ 2 ⋅4 H 2 O , 236.15 g/mol)
needed to prepare 20mL of 1 M Ca(N O 3 ¿ 2 .
*20mL = 0.02 L & 1 M = 1 mol/L
Mass of Ca(N O 3 ¿ 2 = 0.02 L x 1 mol KClO 3 ,
1L x 236.15 g/ mol
1 mol C) Using the dilution formula, calculate the volume of 0.200 M KClO
mL of 0.0100 M KClO 3 .
*M1V1 = M2V2*
M1 = 0.200 M
V1 = M2V2 / M1
M2 = 0.0100 M
V1 = 0.0100 M * 100.0 mL / 0.200 M
V1 =?
V1 = 5mL or 0.005 L
V2 = 100mL = 0.1L 3 needed to prepare 100.0 D) Calculate the mass of solid sodium thiosulfate pentahydrate (Na 2S2O3 ⋅
needed to prepare 250mL of 0.025 M Na2S2O3.
*250mL = 0.25L & 0.025 M = 0.025 mol/L*
Mass of Na2 S2O3 = 0.25 L x 0.025mol Na 2 S 2 O3
1L x ≈ 4.7 g = 4.723 5H2O, 248.18 g/mol) 248.18 g/ mol
1 mol = 1.551125 ≈ 1.6 g
2.
A) Give the chemical equation that describes the equilibrium between calcium iodate and its’ ions
in a saturated soln.
Ca(IO3)2 (s) ⇌ Ca2+(aq) + 2IO3-(aq)
B) Give the chemical equation for the net reaction that occurs during the titration.
IO3-(aq) + 5I-(aq)+ 6H3O+(aq) → 3I2 (aq) + 9H2O(l)
+ 3I2(aq) + 6S2O32-(aq) → 6I-(aq) + 3S4O62-(aq)
------------------------------------------------------------------------Net equation: IO3-(aq) + 6S2O32-(aq) + 6H3O+(aq) → I-(aq) + 3S4O62-(aq) + 9H2O(l) C) How will you know when a stoichiometric amount of the Na2S2O3 soln. has been added in the
titration?
The color of the solution will begin to change as you add the last drop of Na 2S2O3. Once the last
drop is added, the last appearance of blue will disappear, and this will indicate the stoichiometric amount
has been added into the titration.
D) What is the purpose of the trial titration?
The purpose of the trial titration is to acquire and make sure the equivalence point is accurate.
3.
A) From the concentration and volume of KClO 10.0 mL x 1L
1000 mL = 0.01 L KIO 3 x 3 , calculate moles of KClO 0.0100mol
1L 3 present. = 1x10-4 or 0.0001 moles of KIO3 → (IO3-)
B) Using the moles of KClO 3 and the stoichiometry of the balanced net titration reaction,
calculate the moles of Na2S2O3 6 moles S 2 O3 2 – 1 mol IO3
x −¿ 0.0001moles I O 3
¿ – –
= 0.0006 moles Na2S2O3 → ( S 2 O 3 ¿ C) From moles and concentration of Na2S2O3, calculate volume of Na2S2O3 needed to complete the
titration 0.0006 moles N a2 S 2 O3
0.025 = 0.024 L x 1000 mL
1L = 24.0 mL of Na2S2O3 4. What precautions must be observed during this experiment?
Acid is being dealt with which can bring forth a lot of health concerns therefore make sure you wear lab
coat and goggles at all times. Along with wearing the lab coat and goggles, wear gloves and tie any long
hair back from the chemicals throughout the entire lab.
5. Using the SDS, list the health effects of chemical exposure to all the reagent used in this
experiment.
- KIO3: Skin and eye irritation
Ca(NO3)2: Intense fires and damage to organs
Na2S2O3: Respiratory tract irritations and eye irritation
Ca(IO3)2: Respiratory irritation, can cause burns and skin/eye irritation Lab Data
A) 50mL of 0.200 M KIO3
Approximate mass of KIO3 needed: 2.14g
Exact mass of KIO3 weighed: 2.14g
Exact Molarity of KIO3 Soln.: 0.02 M *50mL = 0.05 L 0.01 M
0.05 L = 0.02 M B) 20 mL of ~1 M Ca(NO3)2
Approximate mass of Ca(NO3)2 . 4H2O needed: 4.723g
Actual mass of Ca(NO3)2 . 4H2O weighed: 4.723g
Wt% Ca(NO3)2 in the soln: 19% 4.723 g
4.723 g+ 20 mL x 100 = 19% C) 100 mL of 0.0100 M KIO3
Volume of 0.200 M KIO3 needed:5 mL
Exact molarity of KIO3 soln: 0.02M
*5 mL = 0.005 L* 0.0001mol
0.005 L = 0.02 M D) 250 mL of 0.025 M Na2S2O3
Approximate mass of Na2S2O3 needed: 1.55g
Exact mass of Na2S2O3 weighed: 1.55g
Nominal molarity of Na2S2O3 soln: 0.004 M
*250 mL = 0.25*
0.025 M x 0.25 L
1.55 g = 0.004 Exact Titration Data Tables
Table One
mL KIO3
Mol KIO3
Initial Buret Reading
Final Buret Reading
Volume Na2S2O3
Delivered
M of Na2S2O3
Average M of Na2S2O3 Trial 1
10.00 mL
0.0001M
0.10 mL
24.20 mL
24.10 mL Trial 2
10.00 mL
0.0001 M
24.20 mL
48.35 mL
24.15 mL Trial 3
10.00 mL
0.0001 M
0.10 mL
24.30 mL
24.20 mL 0.0249 M 0.0248 M 0.0248 M 0.0248 M
Calculations:
Trial 1
10.00 mL x 1L
1000 mL x 0.0100mol KIO 3
1L Final Buret reading – Initial buret reading = 0.0001 or 1x10-4 Mol KIO 3 → 24.20-0.10 = 24.10 mL delivered or 0.0241 L 0.0001 M KIO 3 x 2– 1 mol I O 3
1 mol KIO 3 6 mol S 2 O 3
1 mol I O 3 x x 1 mol N a2 S 2 O3
2–
1 mol S2 O3 = 6.00 x 10-4 N a2 S 2 O3 6.00 x 10−4 M N a2 S 2 O3
0.0241 L = 0.024896 Trial 2
Final Buret reading – Initial buret reading 6.00 x 10−4 M N a2 S 2 O3
0.02415 L → 48.35-24.20 = 24.15 mL delivered or 0.02415 L = 0.0248447205 Trial 3
Final Buret reading – Initial buret reading 6.00 x 10−4 M N a2 S 2 O3
0.02420 L ≈ 0.0249 M ≈ 0.0248 M → 24.30-0.10 = 24.20 mL delivered or 0.02420 L = 0.0247933884 ≈ 0.0248 M Na2S2O3 Table Two
Trial Titration Exact Titration Sample 1 Exact Titration
Sample 2 Average M of - Initial Buret Reading
Final Buret Reading
Volume Na2S2O3
delivered
Mol Na2S2O3 delivered
Mol of IO3- Present
M of IO3Both Trials: - 0.10 mL
38.50 mL
38.40 mL 0.20 mL
38.60 mL
38.40 mL - 9.52 x 10-4 Mol
1.59 x 10-4 Mol
0.0159 M 9.52 x 10-4 Mol
1.59 x 10-4 Mol
0.0159 M N a2 S 2 O3 0.0248 M Final Buret reading – Initial buret reading → 38.50-0.10 = 38.40 mL delivered or 0.0384 L Final Buret reading – Initial buret reading → 38.60-0.20 = 38.40 mL delivered or 0.0384 L 0.0384 L x 9.5232 x 10-4 x 0.0248 M N a2 S 2 O3
1L
1 Mol I O 3
6 mol *10 mL = 0.01 L* = 9.5232 x 10-4 Mol = 1.59 x 10-4 Mol IO3 −4 1.59 x 1 0 mol I O3
0.01 L = 0.015872 ≈ 0.0159 M IO3 Post Lab
1. Compare the exact molarity of the Na2S2O3 titrant determined by standardization, to the nominal
molarity of Na2S2O3 calculated based on the mass of Na2S2O3. Should these molarities match? Why
or why not? Try to explain any differences between these two values.
When the nominal molarity was calculated originally, it was 0.004M and when the exact molarity was
calculated by the standardization, it was 0.0248 M. There was an increase in the exact molarity compared
to the nominal molarity. It makes sense that these two molarities would not match because as a reaction
occurs during titrations and mixing, Na2S2O3 would change during that time period thus changing the
molarity calculations/number.
2. Write Molecular, complete ionic and net ionic equations for the precipitation of calcium iodate
from calcium nitrate and potassium iodate.
Molecular: Ca(NO3)2(aq) + 2KIO3(aq) → Ca(IO3)2(S) + KNO3(aq) Complete Ionic: Ca2+(aq) + 2NO32-(aq) + 2K+(aq) + 2IO3-(aq)
Net: Ca2+(aq) + 2IO3(aq) → Ca(IO3)2 (s) + 2K+(aq) + 2NO3-(aq) ⇌ Ca(IO3)2(S) 3.
A) Write the solubility reaction, solubility constant expression and ICE table for the dissolution of
Ca(IO3)2(s).
I
C
E Ca(IO3)2
- ⇌
0
+S
S Ca2+ + 2IO30
+2S
2S B) Which term in the ICE table represents the molar solubility of calcium iodate? What is the
mathematical relationship between molar solubility and Ksp for Ca(IO3)2?
From the table above, the “S” represents the molar solubility of Calcium Iodate. The relationship is
product and reactants → Ksp = [Ca2+][IO3-]2 = (s)(2s)2 = 4s3
Ksp = 4(0.015872)3
Ksp = 1.59939 x 10-5 rounded to 1.60 x 10-5
C) From the molarity of IO3- in saturated soln. #1, determine the equilibrium concentration of Ca 2+
in the soln., determine the molar solubility of Ca(IO3)2 in saturated soln. #1, and the Ksp value.
Repeat the saturated soln. #2
Solution #1
Solution #2
M of IO3
0.0317 M
0.0317 M
M of Ca2+
0.0159 M
0.0159 M Solubility of Ca(IO3)2
Ksp
Calculations:
2s = 2(0.015872)
[IO3-]2 = 0.031744 ≈ 0.0159 M
1.60 x 10-5 0.0159 M
1.60 x 10-5 0.0317 M IO3- D) Compare the two solubilities and the two Ksp values. Are they similar, or very different? Is this
what you had expected? Explain.
The two solubilities and Ksp values are the exact same and it is because their volume measurement taken
is also the same. Although it started and ended at different numbers, their overall volume delivered of
Na2S2O3 was identical meaning all the calculations would also be very similar.
E) Use your textbook or internet resources to look up the Ksp value for Ca(IO3-)2. Calculate the %
error in your experimental result (using the average of your two K sp values) using the formula.
% error = log ( 1.59939 x 10−5 )−log ( 6.47 x 10−6 )
log( 6.47 x 10−6 ) x 100 = 7.57% ...

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- Fall '15
- stellfox
- Solubility, Solubility equilibrium, KClO