PRACTICAL.
Pre-KCSE Practice 1: 2013
1.You are provided with:
(i)solution L containing 5.0g per litre of a dibasic organic acid H2X.2H2O.
(ii)solution M which is acidified potassium manganate(VII)
(iii)solution N a mixture of sodium ethanedioate and ethanedioic acid
(iv)0.1M sodium hydroxide solution P
(v)1.0M sulphuric(VI)
You are required to:
(i)standardize solution M using solution L
(ii)use standardized solution M and solution P to determine the % of sodium ethanedioate in the mixture.
Procedure 1
| Table 1 CT=1/2 mk DP=1/2 mk AC=1/2 mk AV=1 mk FA=1/2 mk Total=3 mk |
Fill the burette with solution M. Pipette 25.0cm3 of solution L into a conical flask. Heat this solution to about 70oC(but not to boil).Titrate the hot solution L with solution M until a permanent pink colour just appears .Shake thoroughly during the titration. Repeat this procedure to complete table 1.
Table 1
| 1 | 2 | 3 | |
| Final burette reading (cm3) | 20.0 | 20.0 | 20.0 |
| Initial burette reading (cm3) | 0.0 | 0.0 | 0.0 |
| Volume of N used (cm3) | 20.0 | 20.0 | 20.0 |
(2marks)
(a)Calculate the average volume of solution L used (1mk)
20.0 + 20.0+ 20.0 = 20.0cm3
3
(b)Given that the concentration of the dibasic acid is 0.05molesdm-3.determine the value of x in the formula H2X.2H2O (H=1.0,O=16.0) (1mark)
Molar mass H2X.2H2O = mass / litre =>
moles / litre
5.0g/litre = 100 g
0.05molesdm-3
H2X.2H2O =100
X = 100 – ((2 x1) + 2 x (2 x1) + (2 x 16) => 100 – 34 = 62
(c) Calculate the number of moles of the dibasic acid H2X.2H2O. (1mark)
Moles = molarity x pipette volume =>
1000
0.05 x 25 = 0.00125 / 1.25 x10 -3 moles
1000
(d)Given the mole ratio manganate(VII)(MnO4–): acid H2X is 2:5, calculate the number of moles of manganate(VII) (MnO4–) in the average titre. (1mark)
Moles H2X = 2/5 moles of MnO4–
=> 2/5 x 0.0125/1.25 x10 -2 moles = 0.0005 / 5.0 x 10 -4 moles
(e)Calculate the concentration of the manganate(VII)(MnO4–) in moles per litre.
(1mark)
Moles per litre/molarity = moles x 1000
average burette volume
=> 0.0005/5.0 x 10 -4moles x 1000 = 0.02083 moles l-1 / M
24.0
Procedure 2
| Table 1 CT=1/2 mk DP=1/2 mk AC=1/2 mk AV=1 mk FA=1/2 mk Total=3 mk |
With solution M still in the burette ,pipette 25.0cm3 of solution N into a conical flask. Heat the conical flask containing solution N to about 70oC.Titrate while hot with solution M. Repeat the experiment to complete table 2.
Table 2 (2marks)
| 1 | 2 | 3 | |
| Final burette reading (cm3) | 12.5 | 12.5 | 12.5 |
| Initial burette reading (cm3) | 0.0 | 0.0 | 0.0 |
| Volume of N used (cm3) | 12.5 | 12.5 | 12.5 |
(a)Calculate the average volume of solution L used (1mk)
12.5 + 12.5 + 12.5 = 12.5 cm3
3
(b)Calculations:
(i)How many moles of manganate(VII)ions are contained in the average volume of solution M used? (1mark)
Moles = molarity of solution M x average burette volume
1000
=> 0.02083 molesl-1/ M x 12.5 = 0.00026 / 2.6 x 10-4 moles
1000
(ii)The reaction between manganate(VII)ions and ethanedioate ions that reacted with is as in the equation:
2MnO4– (aq) + 5C2O42- (aq) + 16H+ (aq) -> 2Mn2+(aq) + 10CO2(g) + 8H2O(l)
Calculate the number of moles of ethanedioate ions that reacted with manganate (VII) ions in the average volume of solution M. (1mark)
From the stoichiometric/ionic equation:
mole ratio MnO4– (aq): C2O42- (aq) = 2:5
=> moles C2O42- = 5/2 moles MnO4–
=> 5/2 x 0.00026 / 2.5 x 10-3 moles
= 0.00065 / 6.5 x10-4 moles
(iii)Calculate the number of moles of ethanedioate ions contained in 250cm3 solution N. (1mark)
25cm3 pipette volume -> 0.00065 /6.5 x10-4 moles
250cm3 -> 0.0065 /6.5 x10-3 moles x 250 = 0.0065 / 6.5 x10-3 moles
25
Procedure 3
| Table 1 CT=1/2 mk DP=1/2 mk AC=1/2 mk AV=1 mk FA=1/2 mk Total=3mk |
Remove solution M from the burette and rinse it with distilled water. Fill the burette with sodium hydroxide solution P. Pipette 25cm3 of solution N into a conical flask and add 2-3 drops of phenolphthalein indicator. Titrate this solution N with solution P from the burette. Repeat the procedure to complete table 3.
Table 3
| 1 | 2 | 3 | |
| Final burette reading (cm3) | 12.5 | 12.5 | 12.5 |
| Initial burette reading (cm3) | 0.0 | 0.0 | 0.0 |
| Volume of N used (cm3) | 12.5 | 12.5 | 12.5 |
(2 mark)
(a)Calculate the average volume of solution L used (1mk)
12.5 + 12.5 + 12.5 = 12.5 cm3
3
(b)Calculations:
(i)How many moles of sodium hydroxide solution P were contained in the average volume? (1mark)
Moles = molarity of solution P x average burette volume
1000
=> 0.1 molesl-1 x 24.9 = 0.00249 / 2.49 x 10-3 moles
1000
(ii)Given that NaOH solution P reacted with the ethanedioate ions from the acid only and the equation for the reaction is:
2NaOH(aq) + H2C2O4 (aq) -> Na2C2O4(g) + 2H2O(l)
Calculate the number of moles of ethanedioic acid that were used in the reaction.(1 mk)
From the stoichiometric equation,mole ratio
NaOH(aq): H2C2O4 (aq) = 2:1
=> moles H2C2O4 = 1/2 moles NaOH
=> 1/2 x 0.00249 / 2.49 x 10-3 moles
=0.001245/1.245 x10-3 moles.
(iii)How many moles of ethanedioic acid were contained in 250cm3 of solution N? (1mark)
25cm3 pipette volume -> 0.001245/1.245 x10-3
250cm3 -> 0.001245/1.245 x10-3 moles x 250 25
= 0.01245/1.245 x10-2 moles
(iii)Determine the % by mass of sodium ethanedioate in the mixture
(H= 1.0,O=16.0,C=12.0 and total mass of mixture =2.0 g in 250cm3 solution) (1mark)
Molar mass H2C2O4 = 90.0g
Mass of H2C2O4 in 250cm3 =
moles in 250cm3 x molar mass H2C2O4
=>0.01245/1.245 x10-2 moles x 90.0
= 1.1205g
% by mass of sodium ethanedioate
=(Mass of mixture – mass of H2C2O4) x 100%
Mass of mixture
=> 2.0 – 1.1205 g = 43.975%
2.0
2. You are provided with 5.0 g solid B. You are to determine the molar mass of solid B.
Procedure
| Table 2 CT=1/2 mk DP=1/2 mk AC=1/2 mk TR=1/2 mk Total=2 mk |
Place 100cm3 of liquid L into a plastic beaker. Determine its temperature and record it at time = 0 in Table 2 below. Stir continuously using the thermometer and record the highest temperature change to the nearest 0.5oC after every 30 seconds. After 120 seconds, add all solid B. Continue stirring and recording the temperature to complete table 2.
Table 2
| Time (seconds) | 0.0 | 30 | 60 | 90 | 120 | 150 | 180 | 210 | 240 | 270 | 300 |
| Temperature(oC) | 20 | 20 | 20 | 20 | 18 | 16 | 14 | 14 | 15 | 16 |
(2mark)
(a)Plot a graph of temperature against time(x-axis)(3marks)
| Graph Scale(plots cover over 1/2 graph paper )=1/2 mk Labelling(both axis)=1/2 mk Plotting all points=1 mk Shape(Extrapolated graph)=1mk Total =3 mk |
(b)From the graph show and determine (2 mark )
(i) the highest temperature change ∆T
∆T = T2 –T1 => 13.4 -20 = 6.6o C
Note ∆T is not – 6.6oC
(ii) the temperature of the mixture at 130 seconds
From extrapolation at 130 seconds = 19.2 oC
(iii)the time when all the solid first dissolved
From extrapolation of the lowest temperature = 220 Seconds
(d) Calculate the heat change for the reaction.(Assume density of liquid L is 1.0gcm-3 ) specific heat capacity is 4.2Jkg-1K-1(1mark)
∆H = mass of liquid L x c x ∆T =>100 x 4.2 x 6.6 = + 2772 J = + 2.772 kJ
1000
(e) Given the molar enthalpy of dissolution of Solid B in liquid L is + 22.176kJ mole-1,determine the number of moles of B used(1mark)
Moles of B = ∆H => + 2.772 kJ = 0.125 moles
∆Hs + 22.176kJ mole-1
(f)Calculate the molar mass of B (1mark)
Molar mass of B = Mass used => 5.0 =>40 g
Moles used 0.125 moles
3(a)You are provided with solid Y. Carry out the following tests and record your observations and inferences in the space provided.
(i) Appearance
Observations inference (1mark)
White crystalline solid Coloured Fe2+ ,Fe3+ , Cu2+ ions absent
(ii)Place about a half spatula full of the solid into a clean dry boiling tube. Heat gently then strongly.
Observations inference (1mark)
Colourless droplets forms on the cooler parts of Hydrated compound/salt
test tube
Solid remain white
(ii)Place all the remaining portion of the solid in a test tube .Add about 10cm3 of distilled water. Shake thoroughly. Divide the mixture into five portions.
Observation Inference (1mark)
Solid dissolves to form a colourless solution Coloured Fe2+ , Fe3+ , Cu2+ ions absent
I. To the first portion add three drops of universal indicator. (1mark)
Observation Inference
pH= 4 weakly acidic solution
II.To the second portion, add three drops of aqueous ammonia then add excess of the alkali.
Observation Inference (1mark)
White ppt, insoluble in excess Al3+ , Pb2+
III.To the third portion, add three drops of sodium sulpide solution.
Observation Inference (1mark)
No black ppt Al3+
IV.To the fourth portion, add three drops of acidified Lead(II)nitrate(IV)solution. Heat to boil
Observation Inference (1mark)
White ppt , persist/remains on boiling SO42-
(b)You are provided with solid P. Carry out the following tests and record your observations and inferences in the space provided.
(i)Place a portion of solid P on a clean metallic spatula and introduce it on a Bunsen flame.
(1/2 mark)
Solid burns with a yellow sooty flame C C // C C bonds
(ii)Add all the remaining solid to about 10cm3 of water in a test tube and shake well. Divide the mixture into 4 portions. (1/2 mark)
Solid dissolves to form a clourless solution Polar organic compound
I. To the 1st portion, test with litmus papers (1/2 mark)
Red litmus paper remain red H+ ions Blue litmus paper turn blue
II. To the 2nd portion, add a little sodium hydrogen carbonate(1/2 mark)
Effervescence/fizzing/bubbles H+ ions
Colourless gas produced
III.To the 3rd portion, and three drops of solution M. Warm(1/2 mark)
Acidified KMnO4 is decolorized R OH, C C // C C bonds// solution M is decolorized IV.To the 4th portion, add three drops of bromine water (1/2marks) Bromine water is decolorizedC C // C C bonds
