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SS3: CHEMISTRY - 1ST TERM

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  1. Volumetric Analysis (Titration) | Week 1
    3 Topics
    |
    1 Quiz
  2. Heat of Neutralization & Redox Titrations | Week 2
    2 Topics
    |
    1 Quiz
  3. Test for Common Gases | Week 3
    1 Topic
  4. Qualitative Analysis I | Week 4
    2 Topics
  5. Qualitative Analysis II | Week 5
    1 Topic
  6. Qualitative Analysis III | Week 6
    2 Topics
  7. Qualitative Analysis IV - Anions | Week 7
    1 Topic
  8. Test for Fat and Oil; Protein; Starch | Week 8
    3 Topics
  9. Petroleum I | Week 9
    4 Topics
    |
    1 Quiz
  10. Petroleum II | Week 10
    4 Topics
    |
    1 Quiz



Lesson Progress
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The heat of neutralization can be determined in the laboratory using a thermally insulated container, with the cork having a thermometer and stirrer.

Method: Transfer 100cm3 of 0.5M hydrochloric acid solution into a beaker. Record the temperature of this solution. Transfer 100cm3 of 0.5M sodium hydroxide into another beaker. Record its temperature also. Allow the two solutions to attain the same temperature. Then transfer the 100 cm3 of sodium hydroxide solution into the hydrochloric acid quickly. Fit the cork immediately. Stir the solution very well. While stirring the solution, take the thermometer reading at time intervals until a constant temperature is obtained. Note the highest temperature obtained.

Results

Initial temperature of the acid and base = T1 oC

Final temperature of the mixed solutions = T2 oC

Change in temperature = T2 – T1 = ΔT

Mass of the mixture solution after neutralization, m = 200g

Calculations

The heat change during the reaction of 100 cm3 of 0.5M HCl = mC ΔT.

The number of moles in 100cm3 of HCl = \( \frac {100 \; \times \; 0.5}{1000} \\ \scriptsize = 0.005\; moles \)

Therefore the enthalpy change during neutralization is heat evolve divide by the number of moles of acid in 100 cm3

= \( \frac{mC \Delta T}{n} \)

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