Determination of the Equilibrium Constant, K sp , for a Chemical Reaction

Determination of the Determination of the Equilibrium Constant, Equilibrium Constant,

KKspsp, for a Chemical , for a Chemical ReactionReaction

By: Bronson WestonBy: Bronson Weston

Background InformationBackground Information

KKspsp is a particular type of equilibrium is a particular type of equilibrium

constant called the solubility product constant called the solubility product constant. constant.

Equilibrium is achieved when an Equilibrium is achieved when an ionic solid dissolves to form a ionic solid dissolves to form a saturated solution. The equilibrium saturated solution. The equilibrium exists between the aqueous ions and exists between the aqueous ions and the precipitate, an undissolved solid. the precipitate, an undissolved solid.

A saturated solution contains the A saturated solution contains the maximum concentration of ions of the maximum concentration of ions of the substance that can dissolve at the substance that can dissolve at the solution's temperature.solution's temperature.

As the concentration of solute, As the concentration of solute, dissolved ions, increases, so does the dissolved ions, increases, so does the rate of reprecipitation. When the rate of rate of reprecipitation. When the rate of reprecipitation equals the rate of reprecipitation equals the rate of dissolution, and there is no more net dissolution, and there is no more net dissolution of solid, equilibrium is dissolution of solid, equilibrium is reached.reached.

KKspsp Equation Equation If given the following reaction:If given the following reaction:

AAnnBBmm(s) (s) n A n Am+m+ (aq) + m B (aq) + m Bn-n-(aq) (aq)

The KThe Kspsp of the reaction is: of the reaction is:

KKspsp = [A = [Am+m+]]nn[B[Bn-n-]]mm

KKspsp = molarity of solution = molarity of solution

solution is saturated, no precipitatesolution is saturated, no precipitate KKspsp< Molarity of Solution< Molarity of Solution

solution is saturated, precipitate is formedsolution is saturated, precipitate is formed KKspsp > Molarity of Solution > Molarity of Solution

solution is unsaturated, no precipitate is formedsolution is unsaturated, no precipitate is formed

MaterialsMaterials• Calcium nitrate, Ca(NOCalcium nitrate, Ca(NO33))22, 0.0900 M, 0.0900 M

• Sodium Hydroxide, NaOH, 0.100 MSodium Hydroxide, NaOH, 0.100 M

• 96- Well Microplate96- Well Microplate

• Beral pipetsBeral pipets

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Procedures- Row 1Procedures- Row 11.1. Arrange the Microplate so that you Arrange the Microplate so that you

have 12 wells across from left to righthave 12 wells across from left to right

2.2. Put 5 drops of water in wells Put 5 drops of water in wells #2 through #12 in the #2 through #12 in the first rowfirst row

3.3. Put 5 drops of .0900 M Ca(NOPut 5 drops of .0900 M Ca(NO33))2 2 in in

well #1 in the first rowwell #1 in the first row

4.4. Add 5 drops of Ca(NOAdd 5 drops of Ca(NO33))22 to well #2 to well #2

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5.5. Using an empty Beral pipet, mix the Using an empty Beral pipet, mix the solution in Well #2 thoroughly by solution in Well #2 thoroughly by drawing the solution into the pipet and drawing the solution into the pipet and then squirting it back several timesthen squirting it back several times

6.6. Calculate the molarity of the solution in Calculate the molarity of the solution in well #2.well #2.

5 drops of 0.0900 M = n moles 5 drops of 0.0900 M = n moles

5 drops5 drops

5 drops Ca(NO5 drops Ca(NO33))22 + 5 drops water = 10 drops + 5 drops water = 10 drops

n moles = 0.0900 M = 0.0450 Mn moles = 0.0900 M = 0.0450 M

10 drops 2 10 drops 2

7.7. Using the empty pipet, draw the solution from well #2 Using the empty pipet, draw the solution from well #2 and put 5 drops into well #3and put 5 drops into well #3

8.8. Put the remaining solution back into well #2Put the remaining solution back into well #29.9. Mix the solution in well #3 with the empty Beral pipet Mix the solution in well #3 with the empty Beral pipet

as before.as before.10.10. Continue this serial dilution procedure, Continue this serial dilution procedure,

adding 5 drops of the previous adding 5 drops of the previous solution to the 5 drops of water in solution to the 5 drops of water in each well down the row until you each well down the row until you fill the last one, well #12. fill the last one, well #12.

11.11. After Mixing the solution in well #12, discard 5 dropsAfter Mixing the solution in well #12, discard 5 drops

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CalculationsCalculations

12.12. Determine the Determine the concentration of concentration of Ca(NOCa(NO33))22 solution solution

in each well, in each well, using the method using the method used in step 6used in step 6

Well #1Well #1 0.0900 M0.0900 M

Well #2Well #2 0.0450 M0.0450 M

Well #3Well #3 0.0225 M0.0225 M

Well #4Well #4 0.0113 M0.0113 M

Well #5Well #5 0.00563 M0.00563 M

Well #6Well #6 0.00281 M0.00281 M

Well #7Well #7 0.00141 M0.00141 M

Well #8Well #8 7.03 x 107.03 x 10-4-4 M M

Well #9Well #9 3.51 x 103.51 x 10-4-4 M M

Well #10Well #10 1.76 x 101.76 x 10-4-4 M M

Well #11Well #11 8.79 x 108.79 x 10-5-5 M M

Well #12Well #12 4.39 x 104.39 x 10-5-5 M M

More ProceduresMore Procedures13.13.Place 5 drops of 0.100 M NaOH in Place 5 drops of 0.100 M NaOH in

each well, #1- #12each well, #1- #12

14.14.Use an empty pipet to mix the solution Use an empty pipet to mix the solution in each wellin each well

15.15.Calculate the concentration of each Calculate the concentration of each reactant, Careactant, Ca+2+2 and OH and OH--, and , and record the data on a record the data on a table.table.

16.16.Allow three or four minutes for Allow three or four minutes for precipitates to form. precipitates to form.

17.17.Observe the pattern of precipitation Observe the pattern of precipitation and record, on the table, which and record, on the table, which solutions form a precipitate.solutions form a precipitate.

I drew this one myself

Well #Well # CaCa+2+2 OHOH-- PrecipitatePrecipitate

Well #1Well #1 0.0450 M0.0450 M 0.0500 M0.0500 M YesYes

Well #2Well #2 0.0225 M0.0225 M 0.0500 M0.0500 M YesYes

Well #3Well #3 0.0113 M0.0113 M 0.0500 M0.0500 M NoNo




Well #7Well #7 7.03 x 107.03 x 10-4-4 M M 0.0500 M0.0500 M NoNo






18.18.Assume that the first solution, the Assume that the first solution, the most concentrated, that does not most concentrated, that does not form a precipitate represents the form a precipitate represents the saturated solution.saturated solution.

19.19.Calculate the KCalculate the Ksp sp of Ca(OH)of Ca(OH)22, using , using

the concentration of Cathe concentration of Ca+2+2 and OH and OH-- ions in the saturated solutionions in the saturated solution

Calculate the limiting reactant:Calculate the limiting reactant:

CaCa+2 +2 + 2OH+ 2OH- - Ca(OH) Ca(OH)22

0.0113 mol Ca0.0113 mol Ca+2+2 x 1.00 mol Ca(OH) x 1.00 mol Ca(OH)22

1.00 mol Ca1.00 mol Ca+2+2) )

= 0.0113 mol Ca(OH)= 0.0113 mol Ca(OH)22

0.0500 mol OH0.0500 mol OH-- x 1.00 mol Ca(OH) x 1.00 mol Ca(OH)22

2.00 mol OH2.00 mol OH--

=0.0250 mol Ca(OH)=0.0250 mol Ca(OH)22

0.0113 mol Ca(OH)0.0113 mol Ca(OH)2 2 < 0.0250 mol Ca(OH)< 0.0250 mol Ca(OH)22

CaCa+2+2 is the limiting reactant is the limiting reactant

Calculate moles of OHCalculate moles of OH- - used in the reaction:used in the reaction:

0.0113 M Ca0.0113 M Ca+2+2 x 2.00 M OH x 2.00 M OH--

1.00 M Ca1.00 M Ca+2+2

= 0.0226 M OH= 0.0226 M OH--

Calculate the KCalculate the Kspsp::

KKspsp = [A = [Am+m+]]nn[B[Bn-n-]]mm

KKspsp = [Ca = [Ca+2+2][OH][OH--]]22

KKspsp = [0.0113 M][0.0226 M] = [0.0113 M][0.0226 M]22

Ksp = 5.77 x 10-6 Ksp = 5.77 x 10-6

Calculate Percent Error:Calculate Percent Error:[(Experimental Value-Actual Value) / Actual Value] x 100[(Experimental Value-Actual Value) / Actual Value] x 100

Actual Value Actual Value K Kspsp = 6.5 x 10 = 6.5 x 10-6 -6

Experimental Value Experimental Value K Kspsp = 5.77 x 10 = 5.77 x 10-6-6

((5.77 x 10((5.77 x 10-6-6)) – (6.5 x 10– (6.5 x 10-6-6)) x 100)) x 100

6.5 x 106.5 x 10-6 -6

= = 11% Error11% Error

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

ExplanationExplanation

ReferencesReferences http://faculty.kutztown.edu/vitz/limsport/http://faculty.kutztown.edu/vitz/limsport/

LabManual/KSPWeb/KSP.htmLabManual/KSPWeb/KSP.htm http://www.jesuitnola.org/upload/clark/http://www.jesuitnola.org/upload/clark/

aplabs.htm#Determination%20of%20theaplabs.htm#Determination%20of%20the%20Solubility%20Product%20of%20an%20Solubility%20Product%20of%20an%20Ionic%20Compound%20Ionic%20Compound

http://mooni.fccj.org/~ethall/2046/ch19/http://mooni.fccj.org/~ethall/2046/ch19/solubility.htmsolubility.htm

Vonderbrink, Sally Ann. Laboratory Vonderbrink, Sally Ann. Laboratory Experiments for Advanced Placement Experiments for Advanced Placement Chemistry Student Edition. Flinn Scientific, Inc. Chemistry Student Edition. Flinn Scientific, Inc. Batavia, IL. 1995.Batavia, IL. 1995.

Determination of the Equilibrium Constant, K sp , for a Chemical Reaction

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