Titration problem answer 2

0.39 g of hydrated sodium carbonate were weighed out on a balance reading to the nearest 0.01 g. This was then dissolved in water and made up to 250 cm³ in a volumetric flask. 25 cm³ portions of this solution were titrated with 0.0100 M hydrochloric acid using methyl orange indicator. The titres were 29.40, 29.35 and 29.30 cm³ . Calculate the number of moles of water of crystallization present per mole of sodium carbonate.

The actual formula on the bottle is given as Na₂CO₃.10H₂O. Evaluate the accuracy and reliability of this experiment (this means error analysis - if you haven't seen this section yet, check it out first).

The equation for the reaction is:

Na₂CO₃(aq) + 2HCl(aq) ⇒ 2NaCl(aq) + H₂O(l) + CO₂(g)

Average titre = (29.40 + 29.35 + 29.30)/3 = 29.35 cm³

\ number of moles of HCl = 0.0100 × 29.35/1000

From the equation Na₂CO₃ º 2HCl

\ number of moles of Na₂CO₃ in 25 cm³= ½ × 0.0100 × 29.35/1000

\ number of moles of Na₂CO₃ in 250 cm³ = ½ × 0.0100 × 29.35/1000 × 10 = 0.00147

Molar mass = 0.39/0.00147 = 266 g mol^-1

Na₂CO₃ = 106 g mol^-1

\ mass of water = 266 - 106 = 160

\ number of moles of water = 160/18 = 8.9

\ formula is Na₂CO₃.9H₂O

Error evaluation

We'll start with all the measurement errors. The tolerances of all the items of equipment can be found in the error analysis section.

Weighing error = 0.39 ± 0.01 g = ± 2.6%

Volumetric flask error = 250 ± 0.24 cm³ = ± 0.10%

Pipette error = 25 ± 0.06 cm³ = ± 0.24%

Burette error = 29.35 ± 0.15 cm³ = ± 0.51%

The burette error is made up of a ± 0.1 cm³ tolerance for a class B burette and a ± 0.05 cm³ error for the reading of the end-point.

This gives a total measurement error of just under 3.5%. The biggest error is the weighing error. using a balance which weighs to the nearest 0.001 g would have reduced the error to 0.26%. These balances are quite expensive (around £800) but are quite common in schools and colleges. The main reason for such a large error here is the small mass used. The same reduction in error could have been achieved by having a 3.9 g sample (and using 0.100 M hydrochloric acid). By using both the bigger sample and the better balance we could have reduced the error to 0.026%. The volume measuring errors could have been reduced by using class A equipment.

We got an answer of 8.9 moles of water of crystallization and then chose the nearest whole number ratio (9). This is within our error analysis limits as 8.9 ± 3.5% = 8.6 to 9.2. However, this range does not include the correct answer of 10 waters of crystallization. Our answer is 1.1/10 × 100 = 11% out. This means there must be some other significant error(s) which we have not accounted for. The most likely error is efflorescence which would reduce the amount of water of crystallization present, particularly at the surface. To reduce this effect we could take our sample from the middle of a new, sealed container.