- You are provided with;
- Solution A, 2M Hydrochloric acid
- Solution B, 0.2M Sodium hydroxide
- 6 pieces of 2cm length of magnesium ribbon.
You are required to determine the mass of magnesium ribbon that reacted with hydrochloric acid.
PROCEDURE I
- Using clean measuring cylinder, measure 50cm3 of solution A into a 100ml glass beaker
- Put one piece of magnesium ribboninto solution A in the 100ml glass beaker and simultaneously start the stop watch
- Record the time taken by magnesium ribbon to get completely finished in the table I.
Repeat procedure (ii) and (ii) using the same solution in procedure (i) adding each piece of solution, M and RETAIN it for procedure II (5marks)
TABLE I
Magnesium ribbon
1st
2nd
3rd
4th
5th
6th
Time taken(s)
(s-1)
time - Plot graph of 1 (vertical axis) against the magnesium ribbon. (3marks)
- From the graph determine the time that would be taken for 5cm piece of the ribbon to get completely finished. (2marks)
PROCEDURE II
Transfer all the solution M from procedure I into a 250ml volumetric flask. Top up the flask to the mark with distilled water and shake. Label as solution N.
- Fill the burette with solution
- Using a pipette and pipette filler, place 25cm3 of solution B in a 250ml conical flask. Add 2 drops of phenolphthalein indicator and titrate with solution
- Record your results in table II. Repeat the titration two more times and complete the table.
TABLE I
I
II
III
Final burette reading
Initial burette reading
Volume of solution N used (cm3)
- Average volume of solution N(1mk)
- Mole of sodium hydroxide, solution B used (1mk)
- Moles of hydrochloric acid, solution N, used. (1mk)
- Moles of hydrochloric acid in 250cm3of solution N. (1mk)
- Moles of hydrochloric acid in 50cm3 of solution A.(1mk)
- Moles of hydrochloric acid in solution A that reacted with all the pieces of magnesium ribbon. (1mk)
- Mass of magnesium ribbon used in the reacted (Mg = 24) (2mks)
- You have provided with solid K carry out the test below and record your observation and inferences in the spaces provided.
- Place all of solid K in a boiling tube. Add 10cm3 of distilled water and shake. Keep the mixture for the test in part (b) below.
- Divide the mixture from (a) above into 4 portions
- To the first portion, add aqueous ammonia drop wise until in excess.
Observation
Inference
½ mk
½ mk
Dip a clean end of glass rod into the second portion, and place in on a non-luminous flame - To the third portion, add four drops of barium chloride solution.
Observation
Inference
½ mk
½ mk
- To the fourth portion, add two drops of acidified potassium manganate (VII) solution
Observation
Inference
½ mk
½ mk
- You are provided with liquid Z. Carry out the tests below.
- Place about 1cm3 of liquid Z on a watch glass and light using a burning splint.
Observation
Inference
½ mk
½ mk
- Place about 5cm3 of liquid Z in a boiling tube. Add 3cm3 of distilled water and shake the mixture.
- Divide the solution above into four portions;
- To the first portion test with litmus papers
Observation
Inference
½ mk
½ mk
- To the 2nd portion, add 2-3 drops of universal indicator.
Observation
Inference
½ mk
½ mk
- To the 3rd portion, add a little Sodium Carbonate
Observation
Inference
½ mk
½ mk
- To the fourth portion add 2-3 drops of acidified Potassium dichromate (VII) solution then warm
Observation
Inference
½ mk
½ mk
- Place about 1cm3 of liquid Z on a watch glass and light using a burning splint.
MARKING SCHEME
-
- TABLE 1
Magnesium ribbon
1st
2nd
3rd
4th
5th
6th
Time Taken (sec)
38.62
42.40
49.78
56.94
62.22
72.56
1/time (s-1)
0.026
0.024
0.020
0.018
0.016
0.014
CT =02
DP=01
ACC=01
T =01
05
Must show on the graph
i/t =21x10-3 √1 1
=47.62sec √1
- TABLE II
i
ii
iii
Final burette reading cm3
19.1
19.0
19.2
Initial burette reading cm3
0.0
0.0
0.0
Volume of solution N used (cm3)
19.1
19.0
19.2
CT=01
DP=01
PA=01
AC=01
FA=01 -
- 1+19.0+19.2√ ½ =57.3 = 19.1cm3 √ ½
3 3 - 0.2 moles x 25= 0.005 moles
1000 - NaOH(aq) + HCl(aq)→NaCl(aq) +H2O(l)
Moles of HCL = Moles of NaOH (mole ratio 1:1)
=0.005moles - moles of HCL in 250cm3 of solution
0.005moles x 250cm3 √ ½ =0.0654 moles√ ½
19.1
Or Answer in (iii)x250cm3
v volume cm3 - Moles of HCL in 50cm3= 2x50 √ ½ =0.1 moles √ ½
1000
Moles of HCL in 50cm3=
2x50 √ ½ =0.1 moles √ ½
1000 - Moles of hydrochloric acid in solution A that reacted with all pieces of Mg ribbon
0.1-0.0654 moles√ ½ = 0.0346 moles √ ½ - mass of magnesium ribbon used (must write equation and show mole ratio) equation √ ½
Mole of magnesium = 0.0346 moles = 0.0173 √ 1
2
0.0173x24= 0.4152g√ ½
- 1+19.0+19.2√ ½ =57.3 = 19.1cm3 √ ½
- TABLE 1
-
-
observation
inferences
Solid dissolves √ ½ to form a clourless solution√ ½
Soluble solid √ ½
Cu2+,Fe2+,Fe3+ present √ ½
-
observations
inferences
burns with a golden yellow flame √ 1
Na+ present √ 1
-
observation
inference
No white ppt
√ 1
Na+,k+,NH4+ Present √ 1
(Accept Ca2+,Mg2+,Pb2+ absent) for √ ½
-
observation
inference
White ppt √ 1
SO32-,SO42-,CO32- Present √ 1
-
observation
inference
purple potassium manganate (VII) decolorized √ 1
SO32-, present √ 1
-
-
-
-
observation
inference
Red and blue litmus papers retain their colour √ ½
Z is neutral √ ½
-
observation
inference
PH=7 √ 1
R-OH present √ 1
-
observations
inference
no effervescence √ ½
RCOOH absent /H+ √ ½
-
observation
inference
colour changes from orange to green √ 1
R-OH Confirmed √ 1
-
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