Chemistry Paper 6 Tips

Mr.Physics

Mr.Physics

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As the title says, I need some help in chemistry paper 6 !! The titration questions prove very tough for me !! I hope somebody would help me !!
 
ThePacifics

ThePacifics

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May June Paper 61 2014 Question number 4.

Let me first explain u what the question says.
It says there are 3 solutions.

A and B are acidic ( Hydrochloric acid) whereas C is a an alkali solution.

Experiment 1:- A burette was filled with solution A of dilute hydrochloric acid to the 0.0 cm3 mark.
O.O cm cube over here means the full burette was filled with dilute HCl. Remember we measure burette from the top. Think it to be like a tap which flows from the bottom and the level decreases. So u see the gap above ( this is the volume lost or poured into solution C ) and then they add 20cm cube of solution C.

The final reading is the one u measure on the diagram. which is 24.4 cm cube ( dont measure it from bottom to up or else u will get 25 point sumthing. MEASURE FROM TOP TO BOTTOM)
The initial is 0 cm cube.
Difference is 24.4

Experiment 2:- Its the same volume of dilute HCL and also same solution C.
the reading will be final reading is 6.1 cm cube. ( again measure from top to bottom)
initial reading will be 0 cm cube. because at first it was full... and then they started pouring into the solution C so the volume decreased.
Difference will be 6.1

c i) neutralisation ( acid and alkali are mixed)
ii) if u remember methyl orange is an indicator. Its orange in neutral, yellow in alkali and red in acid. So when the solution becomes neutral it will become orange and INDICATE to us that the neutralisation has occured.

d) Its simple. So that no other acid remains are left. To clean it up, remove traces of the solution

e) i )Its experiment 1 as we see 24.4 cm cube of dilute HCL was used.
ii) U need to compare like if its more than or less than. greater than or less than like that.
iii) This part is the tricky one. Dilute means more of the water , less of the HCL now in experiment one we used MORE of the DILUTE HCL to neutralise the solution C
but in experiment 2 we used LESS OF THE HCL ... why? becuz it was MORE concentrated than A . B was more concentrated than thats y LESS of it was need to neutralise the solution C.
whereas A was more DILUTE we needed MORE of it to neutralise C.

f) 10 cm cube of solution C means HALF from the ones in A and B so. HALF of the value from experiment 2 that is 6.1 divided by 2. why half? becuz solution C is half so obviously we will need the SAME amount of acid to neutralise it. thats why we will HALVE the acid to get the equal amounts.

g) measuring cylinder is easier and more convenient to use. - advantage
Its not that much accurate:- disadvantge.

h) They ask u to give a method other than titration. Now think of something that could give us a suitable method.

One of it given in the mark scheme is the metal carbonate.
Let me explain it to u

Metal carbonate + acid - Metal chloride + CARBON DIOXIDE.

now we could measure how FAST the carbon dioxide is given off. More the concentration the faster the reaction as more collisions with the HCL molecules and the faster the carbon dioxide given off.

Step 1 - take same volume of both the solutions.Same temperature etc. The ONLY variable we will change is the CONCENTRATION.
Step 2- We will add the suitable reactant. example a metal carbonate. lets take either magnesium or calcium carbonate.
Step 3- we will see effervescne. Now make sure to see the bubbles produced. The FASTER they are produced the more faster the reaction that is in the concentrated solution. There are many ways. U cud count the no.of bubbles given off in a unit time. this one is much better way.
Step 4- we will now compare the results
see how fast the bubbles were given off. ( the volume of bubbles at end of experiment will be the SAME. it will just be faster or slower.)

make sure u FRAME the experiment in a proper step by step way. if u know the method, or the procedure just organise it in a good way.

If u need more help or any other query do ask.
BEST of luck all of u !

I m not 100 percent right always. forgive me if theres sum spelling mistakes or sum slight errors. Do check on internet again if u hav doubt.
 
ThePacifics

ThePacifics

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Its simple
MariaKhan243 said:
Also paper 62 m/j 2014 last question
Click to expand...
u just have to under the steps.

part a ) u just have three steps

Step 1 - Filter the solution. bentonite is INSOLUBLE. once u filter it u will get it as residue.
Step 2- basic experimental process, u need to wash it in distilled water first
Step 3- dry it. ( NOT EVAPORATION. some of its content may react and given out to environment or sumthing like tht m not that sure)
U can dry it with the help of a dryer, hair dryer gently heat it ( maybe)

Parb b) as we hav an alkali and concentration problem.
Simple we will descibe the titration process.

Do u want me to write the titration process?
 
ThePacifics

ThePacifics

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Between

acid + metal carbonate → salt + water + carbon dioxide

u cud try to collect water too but collecting gas is easier. Q.4 2014 Paper 61
 
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