kedah paper 3 2013

KEDAH PAPER 32013

DISCUSSION ON ELECTRICAL CONDUCTIVITY

TITRATIONRUSTING

A student carried out three experiments to investigate the electrical conductivity of three compounds, that is naphthalene, lead(II)bromide and glucose in their molten state.

Diagram 1.1 shows the results obtained from the experiment.

Molten naphthalene

CarbonCarbon

b) Based on the observations in Diagram 1.1, state the inference for these experiments.

Molten lead(II) bromide can conduct electricity while molten naphthalene and glucose cannot conduct electricity.

a)Construct a table to record results for this exp

No. Compound Current, A1 Molten

Naphthalene0.0

2 Molten Lead(II) bromide

0.2

3 Glucose 0.0

c) State operational definition for the electrical conductivity.

• Electrical conductivity shows ammeter needle deflected when the molten lead(II) bromide is electrolysed using carbon electrodes

c) State one hypothesis for this experiment.

Aim : to investigate the electrical conductivity of three compounds, that is naphthalene, lead(II)bromide and glucose in their molten state.

MV : naphthalene, lead(II)bromide and glucose RV : electrical conductivity

c) State one hypothesis for this exp.

ANSWER

lead(II)bromide in molten state can conduct electricity while naphthalene and glucose cannot conduct electricity

MV : naphthalene, lead(II)bromide and glucose RV: electrical conductivity

Aim : to investigate the electrical conductivity of three compounds, that is naphthalene, lead(II)bromide and glucose in their molten state.

d) State all the variables for this exp.

Manipulated variable:naphthalene, lead(II)bromide and glucose

Responding variable

Ammeter reading OR Deflection of needle. Constant variable

Molten state OR Carbon electrodes

Compounds can be classified into ionic compound and covalent compound.Based on the compounds in Diagram 1.1, complete Table 1 by classifying the compounds into ionic or covalent compounds.

Ionic compound Covalent compoundlead(II)bromide Naphthalene and

glucose

2 Diagram 2.1 shows a titration method used to determine the concentration of potassium hydroxide solution.

Acid base indicatorAcid Neutral Alkali

Phenolphthalein

colourless colourless Pink

Methyl orange

red Orange yellow

a)Based on Diagram 2.1, sulphuric acid is added gradually to potassium hydroxide

solution while swirling until the mixture in the conical flask change colour.

State the colour change

ANSWER:Pink to colourless

Alkali to neutral

(b) Name the type of reaction between potassium hydroxide solution and sulphuric acid.

Answer:Neutralisation

(c) Table 2.2 shows the volume of sulphuric acid used in the titration

Calculate the   (i) average volume of sulphuric acid used. Average volume of

sulphuric acid = 10.30 + 10.40 + 10.20 3

= 10.30 cm3

(ii) Calculate concentration of potassium hydroxide

solution

(ii) concentration of potassium hydroxide solution

2KOH (aq) + H2SO4 (aq) K2SO4 (aq) + 2H2O (l)

Mol of sulphuric acid = MV/1000 = 1x 10.3/1000 = 0.0103 molFrom equation, 1 mol of H2SO4 reacts with 2 moles KOHSo, 0.0103 mol H2SO4 0.0206 moles KOH

Mole of KOH = MV/1000 So M= mole x 1000/V = 0.0206 x 1000/25 =0.824 mol dm-3

(ii) concentration of potassium hydroxide solution

Concentration of KOH = 1.0 x 2 x 10.30 25.0 x 1 = 0.82 mol dm-3

(d)The experiment is repeated with 1.0 moldm -3 nitric acid. Predict the average volume of nitric acid needed for the titration with potassium hydroxide solution. 

Predict can either be a)In the form of value, ( if value is used)b)In the form of ‘ more than, less than or no change’ (if value cannot be used)

(d)The experiment is repeated with 1.0 moldm -3 nitric acid. Predict the average vol of nitric acid needed for the titration with potassium hydroxide solution.

Use value! Answer 20.60 cm3

Sulphuric acid is a diprotic acid , contains 2 mol of hydrogen ion per 1 mol of acid Nitric acid is a monoprotic acid contains only 1 mol of hydrogen ion per 1 mol of acid

So if HNO3 is used to replace H2SO4, then the volume is double