(O Level Chem) Acid-base back titration problems

This is a difficult type of acid-base titration problems which requires some practice. When you read such a question, you find there are two reactions making it confusing.

Two reactions in a back titration... 

The first reaction involves the reaction of a reactant A of unknown concentration or unknown percentage of impurity with reactant B of known volume and concentration. Reactant B is present in excess so that all of the reactant A has reacted.

Reactant A, in excess, is added to reactant B... 

To find the amount of unreacted reactant B, a titration with another reactant C of known concentration is performed. Knowing the amount of unreacted B, we can then find the concentration or percentage impurity of reactant A. Such is called a back titration method.

Amount of unreacted A is found by titration... 

Example: 0.324 g of impure sample of sodium carbonate is reacted with 50.00 cm³ of 0.1280 mol/dm³ hydrochloric acid. The excess acid requires 30.10 cm³ of 0.1220 mol/dm³ sodium hydroxide for complete neutralization. Calculate the percentage of sodium carbonate in the sample. (Source: http://answers.yahoo.com/question/index?qid=20090706051656AA9oXPP)

[Solution]

The first reaction involves the reaction between Na₂CO₃ and HCl, while the second is between the excess, unreacted HCl and NaOH. Let's work on the second one by writing down the balanced equation to find the number of moles of the excess, unreacted acid.

Second reaction: HCl(aq) + NaOH(aq) --> NaCl(aq) + H₂O(l)

n_HCl / n_NaOH = 1/1
n_HCl = n_NaOH = c_NaOH x V_NaOH
= 30.0 cm³ x 0.1220 mol/dm³
= 0.0300 dm³ x 0.1220 mol/dm³

The total number of moles of HCl used
= 50.00 cm³ x 0.1280 mol/dm³
= 0.05000 dm³ x 0.1280 mol/dm³

The number of moles of HCl that has reacted with the Na₂CO₃
= total number of moles of HCl used – number of moles of unreacted HCl
= 0.05000 dm³ x 0.1280 mol/dm³ – 0.0300 dm³ x 0.1220 mol/dm³

Now, we can work on the first reaction to find the number of moles of sodium carbonate that is present in the sample.

First reaction: Na₂CO₃(s) + 2HCl(aq) --> 2NaCl(aq) + H₂O(l) + CO₂(g)

n_Na2CO3 / n_HCl = 1/2
n_Na2CO3
= n_HCl/2
= (0.05000 dm³ x 0.1280 mol/dm³ – 0.0300 dm³ x 0.1220 mol/dm³)/2
= 0.00274 mol/2
= 0.00137 mol

Molar mass of Na₂CO₃
= 2(23) + 12 + 3(16)
= 106 g/mol

Mass of Na₂CO₃ present in sample
= 0.00137 mol x 106 g/mol
= 0.145 g

Percentage of Na₂CO₃ present in sample
= 0.145 g/0.324 g x 100 %
= 44.8 %


Here are two more back-titration problems taken from http://www.ausetute.com.au/backtitration.html for your practice. Try it before you refer to the solutions on the web page.

1. 25.00 cm³ of a window cleaner containing ammonia is pipetted into a conical flask. 50.00 cm³ of 0.100 mol/dm³ of hydrochloric acid is added to react with all the ammonia in the conical flask. This is immediately done because the ammonia is volatile. The excess hydrochloric acid requires 21.50 cm³ of 0.050 mol/dm³ sodium carbonate. Calculate the concentration of ammonia in the window cleaner. (Answer: 0.114 mol/dm³)
   
2. 50.00 cm³ of 0.200 mol/dm³ hydrochloric acid is added to a sample of 0.125 g of chalk. The excess hydrochloric acid requires 32.12 cm³ of 0.250 mol/dm³ sodium hydroxide for complete neutralization. Calculate the percentage mass of calcium carbonate in the sample. (Answer: 79.2%) 

(Chinese translations: Acid-base titration 酸碱滴定, Ammonia 氨气, Calcium carbonate 碳酸钙, Concentration 浓度, Excess 过量, Hydrochloric acid 盐酸, Impurity 杂质, Mole 摩尔, Neutralization 中和, Sodium carbonate 纯碱, Sodium hydroxide 氢氧化钠)