Class 12 Chemistry Practicals Volumetric Analysis Estimation Of Sodium Hydroxide

This Class 12 Chemistry practical notes page covers the volumetric analysis experiment "Estimation of Sodium Hydroxide" (Ex. No. 04). It explains the aim, the principle of neutralization with methyl orange and phenolphthalein indicators, the two titrations (standard sodium carbonate against link HCl and link HCl against unknown NaOH), the normality and weight calculations, the result table and the short procedure summary.

Topics Covered

1. Aim of the experiment
2. Principle
3. Procedure: Titration I and Titration II
4. Titration II observations and calculations
5. Weight calculation of NaOH
6. Result
7. Short procedure and Titration I calculations

AIM

To estimate the amount of sodium hydroxide dissolved in 250 ml of the given unknown solution volumetrically. For this you are given a standard solution of sodium carbonate of normality 0.0948 N and hydrochloric acid as link solution.

PRINCIPLE

Neutralization of sodium carbonate by HCl takes place as follows:

Na₂CO₃ + 2HCl → 2NaCl + CO₂ + H₂O

To indicate the end point, methyl orange is used as the indicator.

Neutralization of sodium hydroxide by HCl takes place as follows:

NaOH + HCl → NaCl + H₂O

To indicate the end point, phenolphthalein is used as the indicator.

PROCEDURE

TITRATION - I: (Std Na₂CO₃) Vs (link HCl)

The burette is washed with water, rinsed with link HCl solution and filled with the same HCl solution up to the zero mark. Exactly 20 ml of Na₂CO₃ solution is pipetted out into a clean, washed conical flask. To this solution, 2 to 3 drops of methyl orange indicator are added and the solution is titrated against the HCl link solution from the burette. HCl is added drop by drop till the colour changes from straw yellow to pale pink. The burette reading is noted and the same procedure is repeated to get concordant values.

TITRATION - II: (Link HCl) Vs (Unknown NaOH)

The burette is washed with water, rinsed with link HCl solution and filled with the same HCl solution up to the zero mark. Exactly 20 ml of NaOH solution is pipetted out into a clean, washed conical flask. To this solution, 2 to 3 drops of phenolphthalein indicator are added and the solution is titrated against the HCl link solution from the burette. HCl is added drop by drop till the pink colour disappears completely. The burette reading is noted and the same procedure is repeated to get concordant values.

TITRATION - II OBSERVATIONS

Indicator: Phenolphthalein

CALCULATIONS

Volume of HCl link solution (V₂) = 21.7 ml
Normality of HCl link solution (N₂) = 0.0933 N
Volume of unknown NaOH solution (V₃) = 20 ml
Normality of unknown NaOH solution (N₃) = ?

According to the normality equation: V₂ N₂ = V₃ N₃

\( N_3 = \frac{V_2 N_2}{V_3} = \frac{21.7 \times 0.0933}{20} = 0.1012 \text{ N} \)

Normality of unknown NaOH solution N₃ = 0.1012 N

ESTIMATION: WEIGHT CALCULATION

The amount of NaOH dissolved in one litre of the solution = Equivalent weight x Normality

Equivalent weight of NaOH = 40

The amount of NaOH dissolved in 250 ml of the solution = (Equivalent weight x Normality x 250) / 1000

\( = \frac{40 \times 0.1012 \times 250}{1000} = \frac{4.048 \times 250}{1000} = 1.012 \text{ g} \)

The amount of NaOH dissolved in 250 ml of the solution = 1.012 g

RESULT

1. Normality of HCl link solution = 0.0933 N
2. Normality of unknown NaOH solution = 0.1012 N
3. The amount of NaOH dissolved in 250 ml of the solution = 1.012 g

SHORT PROCEDURE

Normality of hydrochloric acid (HCl) standard solution = 0.1010 N

TITRATION - I (Std HCl) Vs (link NaOH) - SHORT PROCEDURE SET-UP

Indicator: Phenolphthalein

CALCULATIONS

Volume of std HCl solution (V₁) = 20.5 ml
Normality of std HCl solution (N₁) = 0.1010
Volume of NaOH link solution (V₂) = 20 ml
Normality of NaOH link solution (N₂) = ? N

According to the law of normality equation: V₁ N₁ = V₂ N₂

\( N_2 = \frac{V_1 N_1}{V_2} = \frac{20.5 \times 0.1010}{20} = 0.1035 \text{ N} \)

Normality of NaOH link solution (N₂) = 0.1035 N