• Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month
Page
  1. 1
    1
  2. 2
    2
  3. 3
    3
  4. 4
    4
  5. 5
    5
  6. 6
    6
  7. 7
    7
  8. 8
    8
  9. 9
    9
  10. 10
    10

Determine the solubility product of calcium hydroxide

Extracts from this document...

Introduction

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.

Middle

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.

Conclusion

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.

The above preview is unformatted text

This student written piece of work is one of many that can be found in our GCSE Aqueous Chemistry section.

Found what you're looking for?

  • Start learning 29% faster today
  • 150,000+ documents available
  • Just £6.99 a month

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

Not the one? Search for your essay title...
  • Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month

See related essaysSee related essays

Related GCSE Aqueous Chemistry essays

  1. Marked by a teacher

    To determine the amount of ammonia in a sample of household cleaning product, 'cloudy ...

    5 star(s)

    The results of the titration were then used to calculate the concentration of the HCl solution. Part 3: Analysis of 'Cloudy Ammonia': 1. The correct rinsing procedures were used to rinse out all the glassware. (See Table 4.1). 2. A volumetric flask was filled with about two-thirds distilled water.

  2. Marked by a teacher

    My aim for this experiment is to investigate the solubility of salt

    4 star(s)

    My results will be recorded like this. Temp of water...Celsius Expt Mass of salt dissolved 1 2 3 ave Results Temp of water 20 Celsius Expt Mass of salt dissolved 1 5.2 2 5.1 3 5.3 Ave 5.2 Temp of water 30 Celsius Expt Mass of salt dissolved 1 5.7

  1. To investigate the rate of reaction between different concentrations of hydrochloric acid with metal ...

    45ml acid- 5ml water- 1.8 molars (90% concentration) 40ml acid-10ml water-1.6 molars (80 % concentration) 35ml acid-15ml water- 1.4 molars (70% concentration) 30ml acid-20ml water- 1.2 molars (60% concentration) 25ml acid-25ml water- 1 molars (50% concentration) 20ml acid-30ml water -0.8 molars (40% concentration)

  2. Task 1: To determine the polarity of water, hexane and ethanol. Task 2: To ...

    Iodine has covalent bond between atoms and weak intermolecular force between molecules. Iodine doesn't conduct electricity because there is no mobile ions in the solution. Iodine has a low boiling point due to the weak van der Waal's force, so that it is volatile.

  1. Determining the equilibrium constant for the hydrolysis of Ethyl Ethanoate Definition

    General formula of an ester. Ester + Water Acid + Alcohol (reversible) EXPERIMENT DETAILS For full experimental details and for the table of results please refer to handout and attached sheet. EVALUATION Ester + Water Acid + Alcohol Initial a b - - Eqm a-x b-x x x Kc = x2 ' (a-x)

  2. Investigate the factors, which affects how quickly Calcium carbonate reacts with hydrochloric acid.

    The more concentrated the acid the more quickly the reaction will occur. In my prediction I had predicted that the more concentrated the solution the faster the reaction will be .If you look back at the results table and the graph you can see that my prediction was correct.

  1. To investigate the effect of concentration on the temperature rise, heat evolved and heat ...

    Procedure Measure 50 ml of 1 M NaOH using a burette and fix it on a stand. Measure 25 ml of 1 M HCl in a pipette and pour completely into the polystyrene cup. Place this cup under the nozzle of the burette so that when alkali drops from the burette it doesn't spill out.

  2. AIM: To study the effect of solid impurities on boiling point of water and ...

    higher than that of pure solvent The difference between the boiling point of the pure liquid and the boiling point of the solution is âTb = Tb (solution) – Tb (pure) The magnitude of this boiling point elevation is directly proportional to the concentration of solute, expressed as the molality; m.

  • Over 160,000 pieces
    of student written work
  • Annotated by
    experienced teachers
  • Ideas and feedback to
    improve your own work