Which means that for every 1 molecule of H2SO4, 1 molecule of Na2CO3 will react with it.
So to find out the molarity of the acid using 25cm3 of acid to neutralise 0.10 mole of alkali you need to begin by making up an accurate solution of a known molarity, of the alkali.
Na2CO3, has a relative mass of: 23+23+12+16+16+16
= 106g/mol
This total if for 1.0 mol, I want 0.10mol.
So 1M solution= 106g dissolved in 1 dm3
= 1mol/dm3
= 1M
But the volumetric flask = 250cm3
So we need ¼ mole in ¼ of a litre = 1mol dm-3
= 106g mol-1
-------------- = 26.5g required
4
So for 0.10M of solution you need to dissolve 26.5
------ = 2.65g in 250cm3 = 0.1M
10
Now that we have made a accurate solution we can find out the molarity of the acid by using the average result of alkali used.
The average = 28.75cm3
To get the amount of alkali used in dm3 you need to:
28.75cm3
------------- = 0.0287dm3
1000cm3/dm3
Then we need to:
Moles = 0.10mol dm-3 x 0.0286dm3
To get the moles of alkali used in 28.75cm3
So this can be shown as:
Moles = 0.0028moles = 0.0028 moles of acid
1 : 1
Now that we have the molarity of the acid in 28.75cm3
We need to find the molarity of acid in 25.00cm3
So molarity of acid = 0.0028 moles in 25.00cm3
= 1000
----- = 40
25
to get the amount of times 25.00cm3 goes into a litre.
= 40 x 0.0028 moles
= 0.112 mol dm-3
This gives you the exact molarity of the acid in dm3.
From this we can see that the acid was a slightly stronger molarity.
We used 25cm3 of acid : 28.75cm3 of alkali.
It is clearly obvious that the acid is stronger because we used less of it to neutralise more alkali.
So overall out results make sense because the acid was a stronger molarity than the acid.
