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Lesson 1, Topic 1
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# Volumetric Analysis: Acid-Base Titrations

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

### 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. 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 in determining 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;

Example 1

A is a solution of H2SO4(aq) Â containing 2.85 g of the acid in 250cm3 of solution. During a titration experiment, 26.80 cm3 of the acid neutralized 25.00 cm3 of a solution containing 286 gdm-3 of X2CO3.10H2O solution.

Calculate

i. The concentration of the acid in moldm-3

ii. The concentration of the base in moldm-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:Â

Â H2SO4(aq) + X2CO3.10H2O(aq) â†’ X2SO4(aq) + 11H2O(l) + CO2(g)

Solution:

i. 250cm3 contains 2.85 g of the acid

1000cm3 will contain

:- $$\frac{1000 \; \times \; 2.85}{250} \\ = \scriptsize 11.4 gdm^{-3}$$

Molar mass of H2SO4 = (1 x 2) + (32 x 1) + (16 x 4) = 98gmol-1

Concentration in moldm-3 = $$\frac{concentration \; in \; gdm^{-3}}{Molar\;mass} \\ = \scriptsize 11.4 gdm^{-3} \\ = \frac{11.4 gdm^{-3}}{98gmol^{-1}} \\ = \scriptsize 0.166moldm^{-3}$$

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

Therefore, the concentration of the base is 0.124 moldm-3

= 0.124 moldm-3

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

The value of X is:

X2CO3.10H2O = 286

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

2X + 240 = 286

2X = 286 â€“ 240

2X = 46

X = 46/2

X = 23

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

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