Determination%20of%20the%20Equilibrium%20Constant,%20Ksp,%20for%20a%20Chemical%20Reaction

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Determination of the Equilibrium Constant, Ksp,
for a Chemical Reaction

• By Bronson Weston

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Background Information

• Ksp is a particular type of equilibrium constant
  called the solubility product constant.
• Equilibrium is achieved when an ionic solid
  dissolves to form a saturated solution. The
  equilibrium exists between the aqueous ions and
  the precipitate, an undissolved solid.

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• A saturated solution contains the maximum
  concentration of ions of the substance that can
  dissolve at the solution's temperature.
• As the concentration of solute, dissolved ions,
  increases, so does the rate of reprecipitation.
  When the rate of reprecipitation equals the rate
  of dissolution, and there is no more net
  dissolution of solid, equilibrium is reached.

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Ksp Equation

• If given the following reaction
• AnBm(s) ? n Am (aq) m Bn-(aq)
• The Ksp of the reaction is
• Ksp AmnBn-m
• Ksp molarity of solution
• ? solution is saturated, no precipitate
• Ksplt Molarity of Solution
• ? solution is saturated, precipitate is formed
• Ksp gt Molarity of Solution
• solution is unsaturated, no precipitate is formed

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Materials

• Calcium nitrate, Ca(NO3)2, 0.0900 M
• Sodium Hydroxide, NaOH, 0.100 M
• 96- Well Microplate
• Beral pipets

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Procedures- Row 1

1. Arrange the Microplate so that you have 12 wells
   across from left to right
2. Put 5 drops of water in wells 2
   through 12 in the first row
3. Put 5 drops of .0900 M Ca(NO3)2 in well 1 in
   the first row
4. Add 5 drops of Ca(NO3)2 to well 2

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• Using an empty Beral pipet, mix the
  solution in Well 2 thoroughly by drawing
  the solution into the pipet and then squirting it
  back several times
• Calculate the molarity of the solution in well
  2.
• 5 drops of 0.0900 M n moles
• 5 drops
• 5 drops Ca(NO3)2 5 drops water 10 drops
• n moles 0.0900 M 0.0450 M
• 10 drops 2

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1. Using the empty pipet, draw the solution from
   well 2 and put 5 drops into well 3
2. Put the remaining solution back into well 2
3. Mix the solution in well 3 with the empty Beral
   pipet as before.
4. Continue this serial dilution
   procedure, adding 5 drops
   of the previous solution to
   the 5 drops of water in each
   well down the row until
   you fill the last
   one, well 12.
5. After Mixing the solution in well 12, discard 5
   drops

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Calculations
Well 1 0.0900 M
Well 2 0.0450 M
Well 3 0.0225 M
Well 4 0.0113 M
Well 5 0.00563 M
Well 6 0.00281 M
Well 7 0.00141 M
Well 8 7.03 x 10-4 M
Well 9 3.51 x 10-4 M
Well 10 1.76 x 10-4 M
Well 11 8.79 x 10-5 M
Well 12 4.39 x 10-5 M

1. Determine the concentration of Ca(NO3)2 solution
   in each well, using the method used in step 6

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More Procedures

1. Place 5 drops of 0.100 M NaOH in each well, 1-
   12
2. Use an empty pipet to mix the solution in each
   well
3. Calculate the concentration of each reactant,
   Ca2 and OH-, and record the data
   on a table.

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1. Allow three or four minutes for precipitates to
   form.
2. Observe the pattern of precipitation and record,
   on the table, which solutions form a precipitate.

I drew this one myself
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Well Ca2 OH- Precipitate
Well 1 0.0450 M 0.0500 M Yes
Well 2 0.0225 M 0.0500 M Yes
Well 3 0.0113 M 0.0500 M No
Well 4 0.00563 M 0.0500 M No
Well 5 0.00281 M 0.0500 M No
Well 6 0.00141 M 0.0500 M No
Well 7 7.03 x 10-4 M 0.0500 M No
Well 8 3.51 x 10-4 M 0.0500 M No
Well 9 1.76 x 10-4 M 0.0500 M No
Well 10 8.79 x 10-5 M 0.0500 M No
Well 11 4.39 x 10-5 M 0.0500 M No
Well 12 2.20 x 10-5 M 0.0500 M No
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1. Assume that the first solution, the most
   concentrated, that does not form a precipitate
   represents the saturated solution.
2. Calculate the Ksp of Ca(OH)2, using the
   concentration of Ca2 and OH- ions in the
   saturated solution

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Calculate the limiting reactant Ca2 2OH- ? Ca(OH)2 0.0113 mol Ca2 x 1.00 mol Ca(OH)2 1.00 mol Ca2) 0.0113 mol Ca(OH)2 0.0500 mol OH- x 1.00 mol Ca(OH)2 2.00 mol OH- 0.0250 mol Ca(OH)2 0.0113 mol Ca(OH)2 lt 0.0250 mol Ca(OH)2 Ca2 is the limiting reactant
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Calculate moles of OH- used in the reaction 0.0113 M Ca2 x 2.00 M OH- 1.00 M Ca2 0.0226 M OH- Calculate the Ksp Ksp AmnBn-m Ksp Ca2OH-2 Ksp 0.0113 M0.0226 M2 Ksp 5.77 x 10-6
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Calculate Percent Error (Experimental Value-Actual Value) / Actual Value x 100 Actual Value ? Ksp 6.5 x 10-6 Experimental Value ? Ksp 5.77 x 10-6 ((5.77 x 10-6) (6.5 x 10-6)) x 100 6.5 x 10-6 11 Error
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Explanation

• This lab was intended to demonstrate how you can
  figure out a salts solubility constant. When a
  solution is saturated, it is in equilibrium.
  Therefore, when a solution is saturated, we can
  use the current concentrations of the ions to
  determine the Ksp. This is why we used the
  concentrations in the 3rd well to determine the
  solubility constant. We were able to assume that
  the 3rd well was a saturated solution, because it
  was the most concentrated solution that did not
  form a precipitate.

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References

• http//faculty.kutztown.edu/vitz/limsport/LabManua
  l/KSPWeb/KSP.htm
• http//www.jesuitnola.org/upload/clark/aplabs.htm
  Determination20of20the20Solubility20Product20
  of20an20Ionic20Compound
• http//mooni.fccj.org/ethall/2046/ch19/solubility
  .htm
• Vonderbrink, Sally Ann. Laboratory Experiments
  for Advanced Placement Chemistry Student Edition.
  Flinn Scientific, Inc. Batavia, IL. 1995.