Standardization of Hydrochloric Acid with Standard Sodium Carbonate Solution

Standardization of hydrochloric acid with standard sodium carbonate solution relates to the neutralization reaction.

Theory

Sodium carbonate is a primary standard chemical and hydrochloric acid is a secondary standard chemical. Sodium carbonate possesses a high level of purity but contains low reactivity. Sodium carbonate is nontoxic, inexpensive, and available in the market. Sodium carbonate does not absorb moisture from the air quickly (less hygroscopic in nature).

Hydrochloric acid contains less purity and is toxic in nature. It is expensive and not available in the market. It is hygroscopic in nature because it absorbs moisture from surroundings readily (More hygroscopic in nature).

A standard solution is one type of chemical solution whose concentration is accurately known. To prepare this solution, it must dissolve a known mass of solute and dilute the solution to a definite volume. This solution is generally prepared by mixing a definite amount of primary standard substance with a definite volume of distilled water.  

The strength of prepared hydrochloric acid can be known by the method of titration. The hydrochloric acid solution is generally titrated by using a methyl orange indicator against the prepared standard sodium carbonate solution. The color of the solution changes in this titration process due to the formation of the product which is called the endpoint.

Sodium carbonate reacts with hydrochloric acid and forms sodium chloride, carbon dioxide, and water. The endpoint of this reaction is slightly acidic because hydrochloric acid is a strong acid and sodium carbonate is a weak base.  The reaction between sodium carbonate and hydrochloric acid during the standardization of hydrochloric acid with standard sodium carbonate solution is given below:

Na₂CO₃ + 2HCl = 2NaCl + CO₂ + H₂O

A spectator ion is an ion that does not take part in the chemical reaction and is found in solution both before and after the reaction. From the above reaction information, sodium ions act as the spectator ion because they do not take part in this acid-base neutralization reaction. They are found in solution both before and after the reaction.

The net ionic equation shows the type of chemical equation that provides the information only those elements, compounds, and ions that directly take part in the chemical reaction.

The net ionic equation for the above reaction during the standardization of hydrochloric acid with standard sodium carbonate solution is given below:

CO₃^2- (aq) + 2H⁺(aq)  —-> CO₂(g) + H₂O(l)

The amount of hydrochloric acid becomes chemically equivalent to the amount of sodium carbonate present in the reaction. The solution becomes gradually neutral after reacting hydrochloric acid with standard sodium carbonate solution in this acid-base titration reaction.

From the above reaction information, it can be said that 1 mole of Na₂CO₃ reacts with 2 mole of HCl.

1 mole Na₂CO₃ = 2 moles HCl

Apparatus

The required apparatus for the standardization of hydrochloric acid with standard sodium carbonate solution is given below:

1. Conical flask
2. Burette
3. Pipette
4. Funnel
5. Volumetric flask
6. Electric balance
7. Beaker

Reagents and Chemicals

1. Sodium carbonate (Na₂CO₃)
2. Hydrochloric acid (HCl)

Indicator

The required indicator for standardization of hydrochloric acid with standard sodium carbonate solution is methyl orange.

Procedure for Standardization of Hydrochloric Acid with Standard Sodium Carbonate Solution

1. Preparation of standard 0.1M 100 ml Na₂CO₃ solutions: Prepare 100 ml of this standard 0.1M solution by dissolving the calculated amount of anhydrous sodium carbonate in 100 ml distilled water in a 100 ml volumetric flask. It must bear in mind that sodium carbonate should be weighed out very carefully on electric balance.
2. Standardization of supplied HCl solution: Fill a clean and dry burette with the supplied hydrochloric acid fixed to a stand with a clamp. Adjust the volume to a convenient known value by tapping out all the air bubbles at the bottom of the burette and note this in your lab book. Take 10 ml of the prepared 0.1M Na₂CO₃ solution in a 250 ml conical flask, Add 2-3 drops of the methyl orange indicator, and note the color of the solution. Now, titrate by adding acid dropwise from the burette into the solution of the conical flask. This work must continue until the methyl orange becomes a faint yellow and then wash the conical flask by spraying a little distilled water from the wash bottle. Continue titration very carefully until the color becomes orange or faint pink and note in your lab book. Find the volume of consumed acid from initial and final readings. Perform two more titrations calculate the volume of consumed acid and note them in your lab book. Take the average of these three volumes. Now, calculate the strength of the supplied hydrochloric acid.

Table

The table for the Standardization of Hydrochloric Acid with Standard Sodium Carbonate Solution is shown below: (N.B.: All of the following values can be changed during your experiments based on requirements)

Obs. No.	Volume of Na2CO3 taken (mL)	Initial burette reading (mL)	Final burette reading (mL)	Amount of HCl consumed (mL)	Average amount of HCl consumed (mL)
1	10	50	37.2	12.8
2	10	37.2	24.4	12.8	12.8333
3	10	24.4	11.5	12.9

Calculation

The calculation for standardization of Hydrochloric Acid with Standard Sodium Carbonate Solution is given below:

From the balanced neutralization reaction equation, we get the following:

1 mole Na₂CO₃ = 2 moles HCl

We apply the following formula,

V₁S₁/V₂S₂ = ½

Or, S₂= {2 (V1 x S1)/ V₂  }    —————————– (i)

From the above experiment, we get the following values:

The volume of Na₂CO₃ solution, V₁= 10 ml

The concentration of Na₂CO₃ solution, S₁= 0.1M

The volume of HCl solution, V₂= 12.8333 ml

The concentration of HCl solution, S₂=?

From the (i) no equation,

S₂ = {2(10 x 0.1)/12.833}= 0.1558 M

Result: The concentration of supplied hydrochloric acid (HCl) is 0.1558 M.

Standardization-related questions with answer

What is a primary standard substance with examples?

A primary standard substance is a substance that contains higher purity, lower reactivity, higher stability, higher molecular weight, is nontoxic, inexpensive, and absorbs moisture very slowly. Examples of primary standard substances are sodium carbonate (Na₂CO₃), Sodium chloride (NaCl), Oxalic acid (C₂H₂O₄), etc.

What is a secondary standard substance with examples?

A secondary standard substance is a substance that contains lower purity, higher reactivity, lower stability, lower molecular weight, is toxic, is expensive, and absorbs moisture quickly. Examples of secondary standard substances are hydrochloric acid (HCl), Sulphuric acid (H₂SO₄), Sodium hydroxide (NaOH), etc.  

What is a solution in chemistry?

It is a one-type homogeneous mixture consisting of two or more substances. A solution contains a solute and a solvent. The solute in the solution dissolves in the solvent. The specific amount of solute that can be dissolved in a solvent is known as solubility. For example, sodium carbonate when dissolved in distilled water then it is called a solution. In this prepared solution, sodium carbonate acts as a solute, and distilled water acts as a solvent.

What is standardization in chemistry?

Standardization is one type of process that determines the specific concentration of a solution. A prepared standard solution must be required for conducting the standardization process. For this process, the titration method is used mostly.

What is titration in chemistry?

It is a popular technique to determine the concentration of an unknown solution by reacting with a solution of known concentration after adding a suitable indicator. In this process, the titrant whose concentration is known must be added from a burette after adding an indicator to a known quantity of the analyte (the unknown solution) until the reaction is complete.

What is an indicator in chemistry?

An indicator is a type of chemical compound that has the ability to change color in the presence of an acid or base. They are collected from plant pigments and contain mildly acidic or basic character in nature. Examples of indicators are methyl orange, litmus, methyl blue, phenolphthalein, etc.

What is molarity in chemistry?

The molarity of a prepared solution is defined as the total number of moles of solute per liter of solution. It depends on the various physical properties of the system such as pressure and temperature. It is generally represented by M.

Standardization-related mathematical problems with solutions

Problem 1: Calculate the amount of sodium carbonate in order to prepare 300 ml 0.1M solution.

Solution:

The molecular weight of sodium carbonate (Na₂CO₃), M= {(23 x 2)+ (12 x 1)+(16 x 3)} =106 gm

Given that,

The volume of required solution, V=300 ml

The concentration of the required solution, S= 0.1M

The amount of sodium carbonate required for this solution, W=?

We know,

W= SMV/1000

Or, W= {(0.1 x 106 x 300)/1000}

Or, W= 3.18 gm

So, the amount of sodium carbonate in order to prepare 300 ml 0.1M solution is 3.18 gm.

Problem 2: Calculate the amount of sodium hydroxide in order to prepare 500 ml 0.4M solution.

Solution:

The molecular weight of sodium carbonate (NaOH), M= {(23 x 1)+ (16 x 1)+(1 x 1)} =40 gm

Given that,

The volume of required solution, V=500 ml

The concentration of the required solution, S= 0.4M

The amount of sodium carbonate required for this solution, W=?

We know,

W= SMV/1000

Or, W= {(0.4 x 40 x 500)/1000}

Or, W= 8 gm

So, the amount of sodium hydroxide in order to prepare 500 ml 0.4M solution is 8 gm.

Problem 3: 40 ml of an HCl solution is required for neutralization of 20 ml 0.1M Na₂CO₃ solution. Calculate the strength of the HCl acid.

Solution:

The reaction between sodium carbonate and hydrochloric acid is given below:

Na₂CO₃ + 2HCl = 2NaCl + CO₂ + H₂O

From the above reaction information, it can be said that 1 mole of Na₂CO₃ reacts with 2 mole of HCl.

1 mole Na₂CO₃ = 2 moles HCl

Now, we can apply the following formula:

V₁S₁/V₂S₂ = ½

Or, S₂= {2(V1 x S1)/ V₂ }    —————————– (i)

From the above experiment, we get the following values:

The volume of Na₂CO₃ solution, V₁= 20 ml

The concentration of Na₂CO₃ solution, S₁= 0.1M

The volume of HCl solution, V₂= 40 ml

The concentration of HCl solution, S₂=?

From the (i) no equation,

S₂ = {2 (20 x 0.1)/40 } = 0.1 M

Result: The concentration of supplied hydrochloric acid (HCl) is 0.1 M.

Problem 3: 60 ml of an HCl solution is required for neutralization of 50 ml 0.3M Na₂CO₃ solution. Calculate the strength of the HCl acid.

Solution:

The reaction between sodium carbonate and hydrochloric acid is given below:

Na₂CO₃ + 2HCl = 2NaCl + CO₂ + H₂O

From the above reaction information, it can be said that 1 mole of Na₂CO₃ reacts with 2 mole of HCl.

1 mole Na₂CO₃ = 2 moles HCl

Now, we can apply the following formula:

V₁S₁/V₂S₂ = ½

Or, S₂= {2(V1 x S1)/V₂ }    —————————– (i)

From the above experiment, we get the following values:

The volume of Na₂CO₃ solution, V₁= 50 ml

The concentration of Na₂CO₃ solution, S₁= 0.3M

The volume of HCl solution, V₂= 60 ml

The concentration of HCl solution, S₂=?

From the (i) no equation,

S₂ = {2 (50 x 0.3)/ 60 } = 0.5 M

Result: The concentration of supplied hydrochloric acid (HCl) is 0.5 M.