Volumetric Analysis: Acid-Base Titrations

Volumetric Analysis is a method of analysis used to determine the concentration of an unknown substance by titration. Your syllabus requires you to be familiar with acid-base titration and redox titration.

The concentration of a solution is the amount of solute in 1 dm³ or 1000 cm³ of the solution. It is expressed in mol dm^-3 or g dm^-3

Meaning of Titration:

Titration is done by adding a solution (either a standard solution or a non-standard solution) from a burette, slowly to another solution (could either be a standard solution or a non-standard solution) in a conical flask until the reaction is indicated to be complete.

A pipette is used to put an accurate volume of reactant in the conical flask. The pipette used is 20 cm³ or 25 cm³.

Here, the solution (standard or non-standard) from the burette is added to the other solution in a titration flask.

An indicator is used to know when the reaction is complete. This happens by the change of colour by the indicator. The colour change is observed at the endpoint of the titration. The endpoint is the point where the colour change of the indicator is clearly visible.

Application of Acid-Base Titration:

Acid-base titration is applied to determine the following and much more:

1. The standardization of non-standard solutions of acid and alkaline;
2. The molar masses of acids and bases and water of crystallization;
3. The percentage purity and percentage impurity of acids and bases;
4. Stoichiometry of reactions;

Below are common indicators and their colour change in acid and base solutions.

Indicator	Colour in Acid Solution	Colour in Alkali Solution
Litmus	Red	Blue
Methyl orange	Pink	Yellow
Methyl red	Red	Yellow
Phenolphthalein	Colourless	Pink

Example 1.1.1:

A is a solution of H₂SO_4(aq) containing 2.85 g of the acid in 250 cm³ of solution. During a titration experiment, 26.80 cm³ of the acid neutralized 25.00 cm³ of a solution containing 286 g dm^-3 of X₂CO₃.10H₂O solution.

Calculate:

i. The concentration of the acid in mol dm^-3
ii. The concentration of the base in mol dm^-3
iii. The value of X
iv. The percentage by mass of X

[H = 1; S = 32; O = 16; C = 12]

The equation for the reaction is: 

H₂SO_4(aq) + X₂CO₃.10H₂O_(aq) → X₂SO_4(aq) + 11H₂O_(l) + CO_2(g)

Solution:

i. 250 cm³ contains 2.85 g of the acid

1000 cm³ will contain

⇒ \( \frac{1000 \; \times \; 2.85}{250} \\ = \scriptsize 11.4\: g\:dm^{-3} \)

Molar mass of H₂SO₄ = (1 × 2) + (32 × 1) + (16 × 4) = 98 g mol^-1

Concentration in mol dm^-3 = \( \frac{concentration \; in \; gdm^{-3}}{Molar\;mass} \\ = \scriptsize 11.4\: gdm^{-3} \\ = \frac{11.4 \:g\:dm^{-3}}{98\:g\:mol^{-1}} \\ = \scriptsize 0.116\:mol\:dm^{-3} \)

CA = 0.116 mol dm-3	VA = 26.80 cm3	nA = 1
CB = ?	VB = 25.00 cm3	nB = 1

ii. \( \frac{C_A V_A}{C_B V_B} = \frac{n_A}{n_B}\\ \frac{0.116 \; \times \; 26.8}{C_B \; \times \; 25} = \frac{1}{1} \\ \scriptsize C_B = \normalsize \frac{0.116 \; \times \; 26.8 \; \times \; 1}{25 \; \times \; 1}\)

= 0.124 mol dm^-3

Therefore, the concentration of the base is 0.124 mol dm^-3

iii. The mass concentration of X₂CO₃.10H₂O is 286 g/dm³ 

The value of X is:

X₂CO₃.10H₂O = 286

2X + (12 × 1) + (16 × 4) + 10 (1 × 2 + 16) = 286

2X + 240 = 286

2X = 286 – 240

2X = 46

X = \( \frac{46}{2}\)

X = 23

iv. The percentage by mass of X = \( \frac{46}{286} \; \times \; \frac{100}{1} \\ = \scriptsize 16.08 \%\)