Determination of the Solubility Product Lab | Lab Reports Chemistry

Lab # 12 Determination of the Solubility Product:

Purpose: To experimentally determine the Ksp of an ionic compound.

In CHM 111, we classified ionic compounds as soluble or insoluble. In reality, most insoluble ionic compounds dissolve (and ionize) a little in water and are really

slightly soluble or sparingly soluble.

The reaction for the dissolution of a binary ionic compound in water is the reverse of the reaction for the precipitation. The equilibrium constant for the

dissolution of a sparingly soluble ionic compound is called the solubility product (Ksp).

MX(s) ⇌ M+(aq) + X‒(aq) Where Keq = Ksp = [M+(aq)][ X‒(aq)].

The square brackets [ ] imply molar concentration.

Remember that the solid MX(s) is not included in the equation for the equilibrium constant.

This equation works well for ideal solutions (usually low concentrations.) Ksp depends on temperature with the solubility increasing as the temperature

increases (think about sugar in water) for most reactions. There are some ionic compounds whose solubility decreases as the temperature goes up (exothermic

reactions.)

There are often kinetic limits on how fast these reactions occur, so sufficient needs to be allotted for the reaction to come to equilibrium.

We will be looking at the formation of lead(II) iodide to determine the solubility product for PbI2.

PbI2 ⇌ Pb2+ (aq) + 2 I‒ (aq) Ksp = [Pb2+(aq)][ I‒(aq)]2

As with any equilibrium constant, the Ksp will vary with temperature. We will hold the temperature constant in this experiment. In order to get a precipitate,

solutions with variable concentrations of Pb2+ and I‒ will be mixed. If the product of the concentrations of the ions is greater than the solubility product, a

precipitate will form.

The reaction quotient Q = [Pb2+(aq)]i [ I‒(aq)]i2 , where i refers to the concentration of the ion after mixing, can be calculated. The formation of solid crystals of

PbI2 will occur when the value of “Q” exceeds the solubility product value of lead iodide (i.e., when Q > Ksp). Similarly, precipitation will not occur when the

value of Ksp > Q. Thus, in this experimental procedure, if some sample mixtures form PbI2 crystals and other solutions do not, the value of the solubility product

constant lies between Q values with precipitates and Q values without precipitates.

Chemicals: Lead(II) nitrate and KI.

0.010 M Pb2+solution is prepared by dissolving 3.312 grams of Pb(NO3)2 in 1.00 liter of deionized water, while a solution of 0.010 M I‒ is prepared by dissolving

1.660 grams of KI in 1.00 liter of deionized water.

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Lab # 12 Determination of the Solubility Product:

Purpose: To experimentally determine the Ksp of an ionic compound. In CHM 111, we classified ionic compounds as soluble or insoluble. In reality, most insoluble ionic compounds dissolve (and ionize) a little in water and are really slightly soluble or sparingly soluble. The reaction for the dissolution of a binary ionic compound in water is the reverse of the reaction for the precipitation. The equilibrium constant for the dissolution of a sparingly soluble ionic compound is called the solubility product (Ksp). MX(s) ⇌ M+(aq) + X‒(aq) Where Keq = Ksp = [M+(aq)][ X‒(aq)]. The square brackets [ ] imply molar concentration. Remember that the solid MX(s) is not included in the equation for the equilibrium constant. This equation works well for ideal solutions (usually low concentrations.) Ksp depends on temperature with the solubility increasing as the temperature increases (think about sugar in water) for most reactions. There are some ionic compounds whose solubility decreases as the temperature goes up (exothermic reactions.) There are often kinetic limits on how fast these reactions occur, so sufficient needs to be allotted for the reaction to come to equilibrium. We will be looking at the formation of lead(II) iodide to determine the solubility product for PbI 2. PbI 2 ⇌ Pb2+^ (aq) + 2 I‒^ (aq) Ksp = [Pb2+(aq)][ I‒(aq)]^2 As with any equilibrium constant, the Ksp will vary with temperature. We will hold the temperature constant in this experiment. In order to get a precipitate, solutions with variable concentrations of Pb2+^ and I‒^ will be mixed. If the product of the concentrations of the ions is greater than the solubility product, a precipitate will form. The reaction quotient Q = [Pb2+(aq)]i [ I‒(aq)]i^2 , where i refers to the concentration of the ion after mixing, can be calculated. The formation of solid crystals of PbI 2 will occur when the value of “Q” exceeds the solubility product value of lead iodide (i.e., when Q > Ksp). Similarly, precipitation will not occur when the value of Ksp > Q. Thus, in this experimental procedure, if some sample mixtures form PbI 2 crystals and other solutions do not, the value of the solubility product constant lies between Q values with precipitates and Q values without precipitates. Chemicals: Lead(II) nitrate and KI. 0.010 M Pb2+solution is prepared by dissolving 3.312 grams of Pb(NO 3 ) 2 in 1.00 liter of deionized water, while a solution of 0.010 M I‒^ is prepared by dissolving 1.660 grams of KI in 1.00 liter of deionized water.

Materials needed: Test tube rack nine clean, dry test-tubes (15 x 125 mm) 3 10.0 ml pipets 1 pipet bulb Reagent bottles of 0.010 M lead(II) nitrate (aq)and 0.010 M KI(aq) and a bottle of deionized water

Clean and dry 7 medium size test tubes. Label (with pencil) 1 – 7.
Obtain reagent bottles of 0.010 M lead(II) nitrate (aq)and 0.010 M KI(aq) and a bottle of deionized water. Prepare a pipet for each bottle. To prepare, rinse the pipet 3 times with deionized water and then twice with a small amount of the reagent solution. Place the pipet in the reagent bottle for use in the experiment. 3. For runs 1 – 7, pipet the correct volume of each solution into the test tubes. Start by adding the water! Stir when you are done adding all the reactants!!!
Allow the test tubes to sit for at least 30 minutes. Record your results after 30 minutes. 5. Dispose of all the chemical waste in the supplied waste bottle under the hood. The total volume for each solution is 10.0 ml. Note the values for Q decrease from test tube 1 to test tube 9. While you are waiting for the reactions to reach equilibrium (this will take about 30 minutes), calculate Q for reach reaction. The calculations for test tube 1 are below. Test tube

V of DI H 2 O V of 0.010M Pb2+(aq) V of 0.010M I‒ (aq) [Pb2+(aq)]i [ I‒(aq)]i Q = [Pb2+(aq)]I([ I‒(aq)]i)^2 1 1.00 ml 4.00 ml 5.00 ml V 1 M 1 =V 2 M 2 (4.00 ml)(.010M)/(10.0 mL) = M 2 M 2 = 0.004M

V 1 M 1 =V 2 M 2

(5.00 ml)(.010M)/(10.0 mL) = M 2 M 2 = 0.005M

Q =( 0.004M)(0.005M)^2

Q = 1.0x10‒^7 After 30 minutes look for signs of a precipitate in the solution. A golden colored cloudiness indicates that the lead(II) iodide has precipitated. Mixtures of higher Q values will contain shiny, golden crystals of lead iodide, while the tubes of lower Q values will have no solid.

Test tube # V of DI H 2 O V of Pb2+(aq) V of I‒(aq) [Pb2+(aq)]i [ I‒(aq)]i Q = [Pb2+(aq)]I([ I‒(aq)]i)^2 PbI 2 ppt? Appearance of solution in test tube.) 1 1.00 ml 4.00 ml 5.00 ml V 1 M 1 =V 2 M 2 [Pb2+(aq)]i = .004M

V 1 M 1 =V 2 M 2

[ I‒(aq)]i = 0.005M

Q = (.004M)( .005M)^2

Q = 1.0x10‒^7 2 2.00 ml 3.00 ml 5.00 ml 3 3.00 ml 2.00 ml 5.00 ml 4 4.00 ml 1.00 ml 5.00 ml 5 5.50 ml 2.50 ml 2.00 ml 6 6.30 ml 1.20 ml 2.50 ml 7 6.75 ml 1.25 ml 2.00 ml

Add 5 mL of KI(aq) to test tube 7 and stir. Let it sit for a few minutes. Did anything happen? Why?

Looking at the Q values determine the approximate Ksp for PbI 2. Qminimum with PbI 2 (ppt) > Ksp > Qmaximum without PbI 2 (ppt). From your data: Ksp ≅ ____________________________
Look up the actual value for Ksp and calculate the percent error. How close are you to the actual solubility product? What could you have done to improve your results?
Using your approximate value of Ksp, calculate the solubility of PbI 2 in a 0.100 M solution of NaI(aq).

Determination of the Solubility Product Lab, Lab Reports of Chemistry

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Lab # 12 Determination of the Solubility Product:

V 1 M 1 =V 2 M 2

Q =( 0.004M)(0.005M)^2

V 1 M 1 =V 2 M 2

Q = (.004M)( .005M)^2