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

Determine the solubility product of calcium hydroxide

Extracts from this document...


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.

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


    4 star(s)

    base titration and so the theoretical end point is going to be lower therefore an indictor like methyl red which changes colour at 4.2 - 6.3 (from red to yellow)

  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. Preparation of Potassium Trioxalatoferrate (III)

    25cm3 of ethanol was added to the filtrate and any formed crystals were redissolved by gentle heating. The solution was covered with a filter paper and was put in a dark cupboard for crystallisation to occur. The crystals were filtered off and washed with a 1:1 mixture of ethanol and water.

  2. Investigate how the solubility of Potassium Nitrate is affected by Temperature.

    Results Table My Results Temperature Weight Before Weight After Amount of Salt Dissolved 4 oC (10 oC) 31.36g 26.95g 4.41g 20 oC 26.95g 23.04g 3.91g 40 oC 23.04g 18.92g 4.12g 60 oC 18.92g 15g 3.92g 80 oC (85 oC) 15g 10.73g 4.27g Class Results Temperature 1 2 3 4 5

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

    However, hexane and ethanol are miscibility, it is because the dipole moments in ethanol is weak, so that ethanol has the properties of non-polar. Structure of water Structure of iodine In experiment3, iodine is soluble in hexane and ethanol but relatively insoluble in water.

  2. Indirect determination of enthalpy change of decomposition of sodium hydrogen carbonate by thermochemical measurement ...

    I have tried to restrict all calculations to a sensible and meaningful number of significant figures, as a consequence of the possible errors and inaccuracies mentioned above. In addition, given the basic equipment used in this experiment, it would be meaningless for me to calculate a figure for ?H to the nearest thousandth kJ/mol-1, for example.

  1. Determine Solubility of KClO3 Salt.

    Additional solute will not further dissolve if added to a saturated solution. Supersaturated solution is a solution when there is a suitable condition sometimes possible to form a solution that contain greater amount of solute than that needed to form a saturated solution.

  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