INSTRUCTIONS TO CANDIDATES
- Answer ALL questions in this question paper
- Mathematical tables and electronic calculators may be used for calculations.
- All working MUST be clearly shown where necessary.
- You are provided with
- 2.0m NaOH solution labelled B
- H2 SO4 acid solution labelled A `
You are to:- Prepare a dilute solution of NaoH solution.
- Determine the concentration of H2SO4in moles per litre.
PROCEDURE I
- Using a pipette 25.0cm3 of solution B and place it into 250cm3 volumetric flask.
- Add about 200cm3 of distilled water and shake well.
- Add more water to make up to 250cm3mark. Label this solution C.
- Calculate the concentration of the dilute solution C in moles per litres [2mks]
PROCEDURE II
- Fill the burette with solution A and record the readings in the table below.
- Pipette 25cm3 of dilute solution C and place it into 250ml conical flask.
- Add 2 - 3 drops of phenolphthalein indicator.
- Titrate with solution A.
- Record your results in the table below.
- Repeat the titration two or more times and complete the table. [4mks]
Titration no
I
II
III
Final burette reading (cm3)
Initial burette reading(cm3)
Volume of solution A(cm3)
- Determine average volume of the acid (solution A) used. [1mk]
- Determine moles of dilute solution C in the volume used. [2mks]
- Write an equation for the reaction between NaoH and H2SO4 acid. [2mks]
- Determine the number of moles of A (H2SO4) used. [2mks]
- Determine the concentration of A (H2SO4) in moles per litre. [2mks]
- You are provided with the following
- 2M sodium hydroxide solution
- 2M hydrochloric acid
You are required to determine the molar enthalpy of neutralization of the acid using sodium hydroxide.PROCEDURE
- Measure out 20cm3 of acid into a clean plastic beaker.
- Record the temperature of this solution in the table below
- Measure 5cm3 of sodium hydroxide and add it to the hydrochloric acid.
- Stir with the thermometer and record the maximum temperature reached.
- Repeat the above procedure adding 5cm3 portions of sodium hydroxide until the total volume of the solution is 50cm3. [3mks]
Volume of acid(cm3) 20 20 20 20 20 20 20 Volume of NaOH added cm3 0 5 10 15 20 25 30 Temperature(oC)of solution - Plot a graph of temperature rise against sodium hydroxide added. [4mks]
- From your graph
- Determine the expected temperature rise. [2mks]
- Calculate the molar enthalpy of neutralization for this reaction. (C = 4.2J/g/oC) assume density of solution is 1gcm-3)
- The theoretical molar heat of neutralization is -57.2kj/mol-1.Compare your value in [ii] above with the theoretical value. Give the reasons for any differences noted between these two values. [2mks]
- You are provided with solid N carry out the tests below and record your observations and inferences.
- Place a spatular of N in a test tube and add 5cm3 of water and shake well divide the solution in to three portions. [1mk]
OBSERVATION
INFERENCE
- Addsodium hydroxide to the first portion drop wise while observing till in excess. [2mks]
OBSERVATION
INFERENCE
- Add ammonia solution to the second portion drop wise until in excess. [2mks]
OBSERVATION
INFERENCE
- Add four drops of potassium iodide solution to the third portion. [2mks]
OBSERVATION
INFERENCE
- Put spatular of solid P in a test tube and add 5cm3 of water and shake well. Add three drops of acid bariumnitrate followed by 5 drops of nitric acid. [4mks]
OBSERVATION
INFERENCE
- Place a spatular of N in a test tube and add 5cm3 of water and shake well divide the solution in to three portions. [1mk]
MARKING SCHEME
- PROCEDURE I
- 2 moles →1000cm3 2×25= 0.05 moles
? → 25cm3 1000
0.05 moles → 250cm3 1000×0.05 = 0.2M
→1000cm3 250
PROCEDURE II
Complete table 1
Decimal 1
Arithmetic 1
Accuracy 1 ± 0.2- = 12.5cm3
- 0.2 moles → 1000cm3 = 0.005 moles
? → 25cm3 - 2NaoH [aq]+H2SO4[aq]→NaSO4[aq]+2H2O[l]
- Mole ratio
A:C C=0.0025moles
1:2 A= =0.0025 moles - 0.0025 → answer b 1000×0.0025 =_____
→ 1000 answer b
- 2 moles →1000cm3 2×25= 0.05 moles
- Table
Complete table 1
Decimal point 1
Trend 1- Graph
Axis - Mk each
Scale- Mk each
Plotting
Curve -
- answer from the graph
- Use MCΔθ
Mass=50 1=50g.
From MCΔθ→50g J/g/ oC answer in b [i] above. - Theoretical value is higher than the obtained value
Heat lost to the surrounding
Heat absorbed by the apparatus
- Graph
-
-
Observation
Inference
It dissolves into a colourless solution [1mk]
Soluble salt [No Cu2+, Fe2+ or Fe3+] [1mk]
-
Observation
Inference
White precipitate soluble in excess [1mk]
Pb2+, Zn2+, Al3+ ions present[1mk]
-
Observation
Inference
White precipitate insoluble in excess
Pb2+, Al3+ ions present
-
Observation
Inference
Yellow precipitate formed
Pb2+ ions present
-
Observation
Inference
White precipitate [1mk]
SO42- ions
Don’t dissolve on adding nitric acid
Presence of NO, SO32- ions
-
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