New Standard: A.S. 91161 Carry out quantitative analysis

This standard is an internally assessed standard and will involve carrying out an acid-base titration along with chemical calculations involving Moles, Concentration, Volume: n = c x V (V = n ÷ c, c = n ÷ V )

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