B is a solution of Na2CO3. nH2O
prepared by dissolving 5.0g of the hydrated salt in 350cm3 of
solution. A is a solution containing 0.1 moles of hydrochloric acid per dm3
of solution.
a)
Put A into the burette and titrate 20 or 25cm3
portions of B using methyl orange as an indicator. Record the volume of your
pipette. Tabulate your reading and calculate the average volume of acid used.
b)
From your results and information given,
calculate the
i)
Concentration of B in mole per dm3
ii)
Value of Na2CO3. nH2O
iii)
Percentage of water of crystallisation
The equation of reaction is
Na2CO3. nH2O(aq)
+ 2HCl(aq) → 2NaCl(aq) + nH2O(l) + CO2(g)
( C =12, O =16 , Na =23,H=1)
Solution
The volume of the pipette used =
25.00cm3
An indicator used = methyl orange
Titration
|
Burette readings |
Rough |
1st |
2nd |
3rd |
|
Final reading (cm3) |
24.50 |
25.10 |
27.20 |
25.00 |
|
Initial reading (cm3) |
0.00 |
0.00 |
2.10 |
0.00 |
|
The volume of A used (cm3) |
24.50 |
25.10 |
25.10 |
25.00 |
Average volume of A used =
25.1+25.10 + 25.00/3 = 75.2/3 = 25.06cm3
I)
Concentration of B in mol/dm3
CaVa/CbVb = a/b
Ca = 0.1mole/dm3
Va = 25.06cm3
Cb = ??
Vb
= 25.00cm3
.a = 2
.b = 1
0.1
x 25.06/Cbx25 = 2/1
50Cb =
0.1x25.06
Cb = 0.1x25.06/50
= 0.05mol/dm3
II)
value of Na2CO3.nH2O
Mass of anhydrous salt = 0.05 x 106 = 5.3g
Mass of hydrated salt = 5/350 x1000/1 = 14.98g
Mass concentration of hydrated salt/molar mass of hydrated salt = mass of
anhydrous salt/molar mass of anhydrous salt
Mass conc of hydrated salt = 14.The molar mass of hydrated salt = 106
+n18
Mass conc of anhydrous salt = 5.3g
Molar mass of anhydrous salt = 106
14.28/106 +n18 = 5.3/106
5.3(106 + n18) = 14.28 x 106
5.3(106 + n18) = 1513.68
106 + n18 = 1513.68/5.3
106 + n18 = 285.6
.n18 = 285.6 – 106
.n18 = 179.6
.n = 179.6/18
.n =9.98 = Percentagercentage
water of crystallization
Mass of water of crystallization = 14.28 -5.3 =8.98g
Mass of hydrated salt
= 14.28g
%
of crystallization = 8.98/14.28 x 100 = 62.9%
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