Heat of neutralisation (Calculation)

To determine the heat of neutralization between a strong acid (hydrochloric acid) and a strong alkali (sodium hydroxide)

Procedure :

1. Measure 50 cm³ 2.0 mol dm^-3 sodium hydroxide solution and pour it into the polystyrene cup,

2. Put the thermometer into the sodium hydroxide solution. Record the initial temperature,

3. Measure 50 cm³ 2.0 mol dm^-3 hydrochloric acid solution  and record the initial temperature,

4. Pour the hydrochloric acid solution quickly into the sodium hydroxide solution in the

    polystyrene cup.

5. Stir the mixture with the thermometer and record the highest temperature achieved.

Result :

Initial temperature of sodium hydroxide solution =  29.0 ⁰C

Initial temperature of hydrochloric acid  solution =  29.0 ⁰C

Highest temperature of the mixture                     =  42.0 ⁰C

[ Specific heat capacity of solution : 4.2 J g^{-1 0}C^-1 . Density of solution : 1 g cm^-3 ]

Calculate the heat of neutralization.

(Ans: - 54.6 kJ mol^-1)

Heat of combustion (experiment)

Electrolysis (Types of electrodes)

III) Types of electrodes used Electrolysis of 0.2 mol dm-3 copper(II) sulphate solution Using copper electrodes Ions present:
Electrode	Anode (+)	Cathode (-)
Ions attracted
Ion discharged
Half Equation
Observation
Product formed	-
Colour of the electrolyte

Electrolysis of 1.0 mol dm-3 sodium chloride solution

II) The concentration of ions in the solution Electrolysis of 1.0 mol dm-3 sodium chloride solution Ions present:
Electrode	Anode (+)	Cathode (-)
Ions moving
Ion discharged
Reason
Half Equation
Observation
Product formed

Electrolysis of Aqueous solutions

1. Example, sodium chloride solution consists of 4 type of ions

Ions	Cation	Anion
Sodium chloride	Na+	Cl-
Water	H+	OH-

2. Three factors that affect the selective discharge of ions at the

    electrodes during electrolysis of an aqueous solution:

I    The position of ions in the electrochemical series.

II   The concentration of ions in the solution

III Types of electrodes used  

Observation	inference
Colourless gas released	H2 / O2
Greenish yellow gas (Blue litmus → red → white)	Cl2
Brown gas	Br2
Grey solid deposited	Pb/Al
Brown solid deposited	Cu

Electrolysis (molten)

Electrolysis of molten lead(II) bromide Ion present:
Electrode	Anode (+)	Cathode (-)
Ions moving
Half Equation
Observation
Product formed

Answer:

Reactivity of Group 1 elements

When going down Group 1, the alkali metals become more reactive when react with water