Date post: | 18-Jan-2016 |
Category: |
Documents |
Upload: | kristian-lang |
View: | 217 times |
Download: | 0 times |
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