In order to carry out this internal accurately with the aim of achieving at least a merit you will need to be able to carry out an accurate titration-so here goes.....
Watch the 2 videos below to check out how to carry out the titration and then how to go about answering calculations:
Calculations video....
To go along with the above video (from jamesmungall.co.uk) are the written calculations-as set out below:
Titration of Sodium Carbonate against Hydrochloric Acid: ·
2.606g of Anhydrous Na2CO3 was prepared with 250ml of distilled water to make a solution of Sodium Carbonate
· A 25ml aliquot of this solution was titrated against HCL
· Methyl Orange was used as the indicator which changes colour when the Sodium Carbonate is neutralised and turned into Sodium Chloride as per the reaction:Na2Co3+2HCL---------->2NaCl+C02+H2O
· 18.7ml of HCL was required to neutralise the reaction
· WHAT IS THE CONCENTRATION OF THE ACID?
· M(Na2CO3) = 106 gmol-1
n (Na2CO3) = m / M = 2.606 / 106 = 0.0246 Mol
This is the amount of moles in 250ml…we are using aliquots of 25ml
so the amount of moles in 25ml is 0.0246 / 10 = 0.00246 mol
n (HCl) = moles of Sodium Carbonate titred x 2 as the ratio is 1:2*
*from the equation:
Na2CO3 + 2HCl à 2NaCl + Co2 + H20
Mol
ratio: 1 : 2
Therefore Moles of HCl = Moles of Na2Co3 x 2 = 0.00246 x 2 = 0.00492
Therefore the Concentration of HCl required to neutralise the Na2CO3 in this titration is found by using the following equation:
Concentration (HCl) = moles (HCl) / volume (HCl)
C = n / v 0.00492 / 0.0187 = 0.263 Mol/litre
