Need help with this lab

Experiment 12 - Solubility Product.docx

Experiment 12. Solubility Product and Common Ion Effect

Student Name (Print):__________________________

Pre-lab Study Questions (Submit upon arrival for experiment)

1. Write the solubility product K_sp for the following compounds.

a. FeS

b. AgNO₃

c. Mg(OH)₂

d. Ca₃(PO₄)₂

2. Consider a saturated solution of magnesium hydroxide Mg(OH)₂. Answer the following questions.

a. What would happen to the solubility of Mg(OH)₂ if the solution were made up in 0.01 M Mg(NO₃)₂ instead of pure water? Explain.

b. What would happen to the solubility of Mg(OH)₂ if the solution were made up in 0.01 M NaOH instead of pure water? Explain.

c. What would happen to the solubility of Mg(OH)₂ if the solution were made up in 0.01 M NaCl instead of pure water? Explain.

Experiment 12. Solubility Product and Common Ion Effect

Student Name (Print):__________________________

I. Objectives

(1) Measure the solubility of calcium hydroxide Ca(OH)₂ by titration

(2) Determine the common ion effect by measuring the solubility of Ca(OH)₂ with and without the presence of excess Ca²⁺

II. Materials

500-mL beaker

250-mL Erlenmeyer flask

Buret

Solid Ca(OH)₂

1 M HCl

Phenolphthalein indicator

III. Safety Information and Requirements

(1) Wear goggles or safety glasses;

(2) Dump waste according to instruction;

IV. Procedure and Data Collection

IV.1 Information

An ionic compound is formed between a metal cation and a nonmetal anion or a polyatomic ion. When an ionic compound dissolves in water, the compound dissociates into a cation and an anion. For example: NaCl(s) Na⁺(aq) + Cl^-(aq). However not all ionic compounds can completely dissolve in water. Some ionic compounds have low solubility in water and are considered as insoluble compounds. Review solubility rules in General Chemistry I.

Calcium hydroxide Ca(OH)₂ has a low solubility in water. The solubility equilibrium of calcium hydroxide Ca(OH)₂ in water is shown below:

Ca(OH)₂(s) Ca²⁺(aq) + 2 OH^-(aq)

The equilibrium constant of the above reversible equation, aka solubility product of Ca(OH)₂ is given as:

K_sp = [Ca²⁺][OH^-]² (Note that Ca(OH)₂, as a solid, does not appear in the expression of K_sp, in keeping with the rules of equilibrium constant expressions for heterogeneous systems.)

Many factors can affect the solubility of a solid, including the presence of other ions in solution. If an ion present in the solution in which the solid is dissolved is also one of the ions from the solid, this ion is called a common ion. The presence of a common ion shifts the equilibrium to the reactant side of the solubility equilibrium, i.e. the common ion decreases the solubility of the solid. This is known as the Common Ion Effect. In this experiment, we will examine the solubility of calcium hydroxide Ca(OH)₂, a slightly soluble solid and the common ion effect.

IV. 2 Procedure

Preparation of Saturated Ca(OH)₂ Solutions

(1) Measure 2 grams of solid Ca(OH)₂ to a 250-mL Erlenmeyer flask. Fill the flask about half full with 0.010 M Ba(NO₃)₂ solution.

(2) Close the flask with a cork or a stopper. Shake the solution for 5 minutes. Label the solution as Ca(OH)₂ Solution 1.

(3) Measure 2 grams of solid Ca(OH)₂ to another Erlenmeyer flask, repeat steps 1-2 using 0.010 M Ca(NO₃)₂ solution. Label the solution as Ca(OH)₂ Solution 2.

(4) Filter the above two solutions using filter paper. Label the two filtered solutions as Saturated Ca(OH)₂ Solution 1 and Saturated Ca(OH)₂ Solution 2.

Titration of the Ca(OH)₂ Solutions

(1) Record the molarity of the standardized HCl solution provided by the instructor.

(2) Use a pipet to transfer 25.00 mL Saturated Ca(OH)₂ Solution 1 to a clean Erlenmeyer flask. Add 2-3 drops of the indicator to the flask.

(3) Titrate the 25.00 mL of Saturated Ca(OH)₂ Solution 1 with the HCl solution to the end-point, twice. Record the mL of the HCl solution used in titrations. Take the average of the volumes and convert the average volume to L.

(4) Repeat Steps (2) and (3) with Saturated Ca(OH)₂ Solution 2. Record the mL of the HCl solution used in titrations. Take the average of the volumes and convert the average volume to L.

Think:

Ca(OH)₂ has low solubility in water. In which solution, Ba(NO₃)₂ or Ca(NO₃)₂, do you think Ca(OH)₂ has lower solubility than in pure water? (Hint: which solution has a common ion? How does a common ion affect the solubility of an insoluble compound?)

Record

Molarity of the standard HCl solution

M_HCl = _________________M

	Saturated Ca(OH)2 Solution 1		Saturated Ca(OH)2 Solution 2
	Trial 1	Trial 2	Trial 1	Trial 2
VHCl (mL)
Average VHCl (L)

V. Calculation and Conclusions

V. 1 Information

In this experiment, acid HCl is used to titrate base Ca(OH)₂. With the known molarity M of HCl and the measured volume V of HCl to titrate 25.00 mL Ca(OH)₂, we can calculate the molarity M of Ca(OH)₂. Since 1 mol Ca(OH)₂ dissociating in water produces 1 mol Ca²⁺, the calculated molarity M of Ca(OH)₂ is equal to the molarity of Ca²⁺ in Saturated Ca(OH)₂ Solution 1 where no common ion is present:

Ca²⁺ in Saturated Ca(OH)₂ Solution 1 = M of Ca(OH)₂ calculated from the titration equation

However, when the saturated solution is prepared in 0.01 M Ca(NO₃)₂, the molarity of Ca²⁺ is equal to the sum of Ca²⁺ from Ca(OH)₂ and the Ca²⁺ from 0.01 M Ca(NO₃)₂, which is 0.01 M:

Ca²⁺ in Saturated Ca(OH)₂ Solution 2 = M of Ca(OH)₂ calculated from the titration equation + 0.01 M

V.2 Calculation

Plug the average volumes of the HCl solution from Data Collection into the titration equation to calculate the molarities of Ca²⁺ in the two saturated solutions. Complete the following table.

Saturated Ca(OH)2 Solution 1
Average HCl volume VHCl (L)
Moles of H+ (mol) = MHCl VHCl
[OH-] = Moles of H+/0.025 (M)
[Ca2+] = 2 [OH-] (M)
Ksp = [Ca2+][OH-]2
Saturated Ca(OH)2 Solution 1
Average HCl volume VHCl (L)
Moles of H+ (mol) = MHCl VHCl
[OH-] = Moles of H+/0.025 (M)
[Ca2+] = 2 [OH-] + 0.01 (M)
Ksp = [Ca2+][OH-]2

V.3 Conclusions

Compare the [OH^-] in two saturated solutions, which one is smaller? What can you conclude?

Compare the two K_sp values, are they close? What can you conclude?