# MOLAR SOLUBILITY AND COMMON ION EFFECT Assignment | Online Assignment

Watch the following YouTube videos for this experiment: https://www.youtube.com/watch?v=0N174A4Hhcs https://www.youtube.com/watch?v=54rkHFnIQk0 For your lab report: Model your procedure based on the experiment you watched in the video. Create a table for the data presented in the experiment. Report your values for solubility of Ca(OH)2 for both sets of data in units of M and g/100 mL. Also report the experimental value of Ksp obtained from your data. Perform the appropriate error analysis and comment on the differences you observe between the two solutions. Attached is a sample lab report of how it should be formatted.

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PRINCIPLES OF CHEMISTRY

EXPERIMENT #7:  MOLAR SOLUBILITY AND COMMON ION EFFECT

Objectives

• To determine the Ksp of Ca(OH)2
• To investigate the Common Ion effect

Guiding Questions

• Compare your results to the true value of Ksp = 5.5×10-6 for Ca(OH)2. How do you evaluate the accuracy of your measurements?

Introduction

Calcium hydroxide, Ca(OH)2, also called slaked lime, is an ionic compound with limited solubility in water. When mixing Ca(OH)2 with water, only a small proportion dissolves.  We refer to this as a saturated solution.  The dissolution reaches equilibrium when the rate of the backward reaction equals the rate of the forward reaction (Equation 1). A saturated solution of Ca(OH)2 produces a dynamic equilibrium between Ca(OH)2(s) and its ions, Ca2+ and OH (Equation 1).

Ca(OH)2(s)   Ca2+(aq) + 2 OH(aq)                                                                 (1)

This equilibrium lies far to the left because of the low solubility of Ca(OH)2.  The equilibrium constant for Equation 1 is called the solubility product constant (Ksp) of Ca(OH)2, given by Equation 2.

Ksp = [Ca2+]∙[OH]2                                                                                                     (2)

The concentration of Ca(OH)2 does not appear in the equilibrium constant expression because it is present as the pure solid, no matter how much or how little of it is present.

In pure water, each unit of Ca(OH)2 dissociates to give one Ca+2 and two OH ions.  Thus the solubility of Ca(OH)2, which we call s, is equal to the concentration of Ca2+ ions or ½ [OH] ions.

s = [Ca2+]  = ½ [OH]  (in pure water)                                                                        (3)

Since s is a concentration, it is often reported in units of molarity (M), moles solute per liter solution.  However, often it is useful to know the solubility in units of grams solute per 100 mL solution (g/100 mL).  You will be asked to report both in this lab.

The small Ksp value for Ca(OH)2 (5.5×10-6) means that in pure water the molar concentration of Ca2+ is also small, and can be calculated according to Equations 2 and 3:

Ksp = [Ca2+][OH]2 = ½ [OH]3 = 5.5×10-6                 [OH] = 0.022 M                     (4)

According to Equation 4, the molar solubility of Ca(OH)2 = [Ca2+] = ½ [OH] = 0.011 M

Le Chatelier’s Principle tells us that the addition of either calcium or hydroxide ions to a saturated solution will shift the equilibrium to the left, decreasing the solubility of calcium hydroxide.  This shift is known as the common ion effect.  In this lab, you will explore the common ion effect by determining the solubility of calcium hydroxide in an aqueous solution of calcium chloride.  This will be accomplished by titrating the solution with HCl.

Watch the following YouTube videos for this experiment:

Model your procedure based on the experiment you watched in the video. Create a table for the data presented in the experiment. Report your values for solubility of Ca(OH)2 for both sets of data in units of M and g/100 mL. Also report the experimental value of Ksp obtained from your data. Perform the appropriate error analysis and comment on the differences you observe between the two solutions.

# Sample Laboratory Report

Principles of Chemistry 1

Determination of percent composition of KClO4

September 8, 2019

Objectives:  To determine the percent oxygen in KClO3 and to evaluate the precision and accuracy of the measurement.

Introduction: This lab will demonstrate the decomposition (breakdown) of KClO3 into KCl solid and O2 gas when heated. The oxygen will be released into the atmosphere and the remaining KCl can be measured to determine the change in mass. Using the Law of Conservation of Mass we can determine the amount of oxygen released and calculate the percent of oxygen present in the original sample. The chemical reaction we will study is shown below:

2KClO3  → 2KCl(s)  +  3O2(g)

Procedures:

A sample of KClO3 weighing approximately 1 gram was obtained.  After firing a crucible and lid for 10 minutes to remove all impurities, it was allowed to cool to room temperature and weighed with added KClO3 sample.  After heating for 15 minutes, and cooling to room temperature, the crucible, lid and contents were weighed.  The product was heated again for 5 minutes, cooled and re-weighed to see if more oxygen was lost.  This procedure was repeated on 2 additional samples of KClO3.

Safety:

Care was taken to handle the crucible with tongs and to avoid keeping any flammable materials near the Bunsen burner flame.  The waste KCl was not considered hazardous and was disposed of in the trash.

 Results:   (1) Mass of empty dish 42.56 g Mass of dish + KClO3 43.66 g Mass of dish + KCl (after heating) Mass of dish + KCl (reheated)   (2) Mass of empty dish Mass of dish + KClO3 Mass of dish + KCl (after heating) Mass of dish + KCl (reheated)   (3) Mass of empty dish Mass of dish + KClO3 Mass of dish + KCl (after heating) Mass of dish + KCl (reheated) 43.30 g 43.25 g   44.63 g 45.66 g 45.29 g 45.28 g   40.72 g 41.93 g 41.51 g 41.51 g

Sample Calculations:

• Mass of KClO3: 43.66 g -42.56 g = 1.10 g
• Mass of KCl: 43.25 g -42.56 g = 0.69 g
• Mass of O lost: 1.10 g – 0.69 g = 0.41 g
• Experimental % of oxygen using data:
• Actual % of oxygen using periodic table:
• % error

Results Table

 Mass of KClO3 (g) Mass of O (g) % oxygen 1 1.10 0.41 37 2 1.03 0.38 37 3 1.21 0.42 35

Average % O =

Standard Deviation = 0.94 %

% Error =

Discussion:

The Law of Conservation of Mass means that matter cannot be created or destroyed.   Therefore, when the sample is heated, the difference between the original and remaining mass must be due to the mass of oxygen.  Our experiment showed that the percent oxygen in the KClO3 was 36 ± 0.94%.  The standard deviation shows that the % could have been as high as 37% or as low as 35% based on the measurements that were made.  Since the true value is 39%, larger than the experimental error, our values were lower than expected.  This could due to not heating the sample enough to drive off all the oxygen – this would cause the % to be too low.

Conclusion: Decomposition is a type of chemical reaction in which a single reactant breaks apart to form 2 or more new substances. When any chemical reaction occurs, the total mass of the reactants must equal the total mass of the products. This is the Law of Conservation of Mass. We can use this to calculate mass that is either lost or gained during a chemical reaction.  In this experiment, our % oxygen was 37 ± 0.94 %, lower than the actual % oxygen in the formula (39%). This means that we probably did not heat the dish long enough for all of the KClO3 to fully decompose and that there was some KClO3 left after we heated it.

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