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Determine the solubility product of calcium hydroxide

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BACKGROUND Calcium Hydroxide, Ca (OH) 2 is an ionic solid that is slightly soluble in water. A saturated solution of a sparingly soluble salt obeys the Law of Chemical Equilibrium. Therefore, Ca (OH) 2 (s) - Ca2+ (aq) + 2OH-(aq) Keq = [Ca2+ (aq)] [OH-(aq)] 2 [Ca (OH) 2(s)] Keq is the Equilibrium Constant of the reaction. Whenever you see the symbol Ksp you know that it is referring to a solubility equation, written with the solid to the left of the equilibrium sign, and the dissolved products to the right. Since the concentration of the solid Ca (OH) 2 is a constant, it maybe included in the Keq for the reaction, and a new constant Ksp, the Solubility Product, is obtained. Thus for Calcium Hydroxide: Ksp = [Ca2+ (aq)] [OH-(aq)] 2 Every substance that forms a saturated solution will have a Ksp. However, for very soluble substances like NaCl, the value is so large that the concept is rarely used. In slight and low solubility substances, the value of Ksp is a useful quantity that lets us predict and calculate solubility of substances in solution. Ksp is also known as the course of the equilibrium constant, and is constant at constant temperature. ...read more.


For the solutions II, III, and IV, calculate the concentration of Hydroxide ion which is derived from the dissolved Calcium Hydroxide. This is done by subtracting the Hydroxide ion concentration derived from the Sodium Hydroxide from the total Hydroxide ion concentration. Hence, calculate the calcium ion concentration in each of the solutions I, II, III, and IV. Solution II M2 = [OH-] = 0.0913M Contribution from Noah (0.1M) [OH-]Ca (OH) = 0.0913M - 0.1M = -8.7 x 10-3 M Solution III M2 = [OH-] = 0.0562M Contribution from NaOH (0.05M) [OH-]Ca (OH) = 0.0562M - 0.05M = 0.006 M Solution IV M2 = [OH-] = 0.0432M Contribution from NaOH (0.025M) [OH-]Ca (OH) = 0.0432M - 0.025M = 0.0182 M Calculate the calcium ion concentration in each of the solution I, II, III, IV. Ca (OH) 2(s) Ca2+ (aq) + 2OH- (aq) 1 mol of Ca2+ = 2 mol of OH- Solution I [OH-] = 0.0325M [Ca2+] = 0.0325 = 1.62 x 10-2 M 2 Solution II [OH-] = -8.7 x 10-3 M [Ca2+] = -8.7 x 10-3 = -4.35x 10-3 M 2 Solution III [OH-] = 0.006 M [Ca2+] = 0.006 = 3.00 x 10-3 M 2 Solution IV [OH-] = 0.0182 M [Ca2+] = 0.0182 = 9.10 x 10-3 M 2 3) ...read more.


Besides that, there may also be the possibility that, while conducting the experiment, some human error may have occurred or mistakes made by me. Firstly, the parallax error, while reading the burette or pipette, the meniscus of the solution may not be exactly on the line of the measurement. Thus, the measurement may be slightly out Furthermore, the apparatus that we used, maybe clean properly thus this will also affect the final results. As the solutions are left in open air, there is a high possibility that the solutions are contaminated with some impurities and affect the concentration of the solution. While conducting this experiment, there are a few pre-caution that we must take to rest assured the experiment goes and runs smoothly. Firstly, we must always wear our lab coat as we are dealing with harmful acid and bases. We would not wan the solutions to still on our clothes. Second, we should listen carefully and follow exactly the instructions given by the lecturer for the experiment. We must pay full attention in the lab and must not joke around to avoid any mistake or accidents. CONCLUSION The Ksp value for: 1) Solution 1 = 1.81 x 10-5 2) Solution 2 = -3.29 x 10-4 3) Solution 3 = 1.08 x 10-7 4) Solution 4 = 3.014 x 10-6 REFERENCE Books: Chemistry 7th edition - Raymond Chang Websites: 1. http://www.csun.edu/chemistry/Ksp.pdf 2. http://www.webster-dictionary.org/definitin/Calcium%20hydroxide 3. http://www.tarleton.edu/~alow/1084exp6.htm 4. http://www.sasked.gov.sk.ca ...read more.

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Here's what a teacher thought of this essay

3 star(s)

This experimental report was well documented and a series of complex calculations carried out correctly. The values obtained should have been compared with expected values by using references. There was detailed background information in the report.
The evaluation of the experiment was a little weak. The candidate should have looked for alternative explanations as to why there was a negative value obtained for the second solution .A saturated solution of calcium hydroxide must be made fresh on the day it is to be used as any carbon dioxide that enters the solution will cause it to react to form a calcium carbonate precipitate. The candidate did not keep the solutions covered and did not mention this possible source of error.
Temperature will also affect the Ksp value and this was not recorded or mentioned in the analysis.
The candidate would lose marks for reporting Ksp without units on all occasions.

Marked by teacher Stevie Fleming 01/01/1970

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