[Chemistry Class Notes] on Preparation of Standard Solution of Oxalic Acid Pdf for Exam

Oxalic acid is an organic compound that has the molecular formula HO₂C-CO₂H and bears the IUPAC name ethanedioic acid. It is considered the simplest dicarboxylic acid. It is a white crystalline solid that forms a colourless solution when dissolved in water. As the early researchers isolated the oxalic acid from the flowering plant whose genus was named Oxalis, commonly known as wood-sorrels, the compound acquired its name as oxalic acid.  

Though it happens to occur naturally in many food items, prolonged contact with the skin or excessive ingestion of oxalic acid can be very dangerous to human health. Oxalic acid is a reducing agent and acquired a strength that is much larger than acetic acid. Its conjugate base is known as oxalate and is also a chelating agent for metal cations. Oxalic acid is mostly seen to be naturally occurring in the form of dihydrate with the formula C₂H₂O₄・2H₂O

Process of Preparation

The standard solution of oxalic acid is a known high purity substance that can be dissolved to produce a primary standard solution in a known volume of solvent. To prepare a certain quantity of oxalic acid, the respective known solvent weight is dissolved. It is ready using a standard, like a primary standard substance. Let us look at the preparation of standard oxalic acid solution briefly.

Aim

To prepare a standard solution of oxalic acid of M/10 or to prepare the standard solution of 0.1 m oxalic acid.

Theory

Hydrated oxalic acid => C₂H₂O₄.2H₂O

The molecular mass of Oxalic Acid => 126.

 

=> 12.6 g of oxalic acid/litre of the solution should be dissolved in order to produce an M/10 oxalic acid solution.

On the other side,

12.6 /4 = 3.15 g of crystals of oxalic acid should be dissolved in water, and 250 ml of the solution should be produced precisely.

Required Materials

1. Weighing tube

2. 250ml measuring flask

3. Watch glass

4. 250ml beaker

5. Glass rod

6. Chemical balance

7. Wash bottle

8. Weight box

9. Oxalic acid

10. Funnel

11. Funnel stand

12. Distilled water
    

Apparatus Setup

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Procedure

1. Take a watch glass and wash it with distilled water and later dry it.

2. Weigh the exact amount of dried and clean watch glass and record its weight in the notebook.

3. Weigh accurately on the watch glass 3.15 g of oxalic acid and note this weight in the notebook.

4. Transfer oxalic acid softly and carefully using a funnel from the watch glass into a clean and dry measuring flask.

5. Then, wash the watch glass using distilled water to move the particle position that sticks to it into the foam using the assistance of a wash bottle.

6. For this reason, the volume of distilled water should not exceed more than 50 ml.

7. Wash funnel as often as distilled water to move the position of the sticking particle into the measuring flask with a wash bottle. Add water in little quantities while washing the funnel. The distilled water quantity being used for this purpose should not exceed 50 ml.

8. By using a wash bottle, carefully wash the funnel with distilled water to pass the solution attached to the funnel into the measuring flask.

9. Turn the measurement flask until the oxalic acid gets dissolved.

10. Again, using a wash bottle, add enough distilled water thoroughly to the measuring flask just below the etched mark on it.

11. After that, add the remaining few ml of distilled water drop into the measuring flask until the reduced meniscus level touches the mark.

12. Keep the stopper on the mouth of the flask and shake softly to maintain uniformity in the entire solution. Now, calculate it as a solution of oxalic acid M/10.
    

Observations

 

 

Results

250cm³ of (M/10) solution or decimolar of oxalic acid is prepared.

Precautions

1. The oxalic acid crystals need weights of 2g + 1g + 100mg + 5mg while weighing.

2. Wash the watch glass carefully, ensuring that not even a single crystal of oxalic acid is left on the watch glass.

3. Using a pipette, add the remaining few drops to avoid an extra addition of distilled water to the mark above the neck of the measuring cylinder.

4. If it is needed to titrate oxalic acid or oxalate, add the necessary dilute amount of H₂SO₄ and heat the flask at a range of 60° – 70° C.
   

Preparation of Acid from the Desired Normality or Molarity from Its Concentrated Solution

Many acids such as Hydrochloric acid (HCl), Nitric acid (HNO₃), Sulfuric acid (H₂SO₄) and Acetic acid (CH₃COOH) are available in the concentrated form. As they are needed to be used in diluted form, thus, they are mixed with water to acquire the desired volume of diluted acid.  If the concentration of the concentrated acid is known, it gets very easy to determine the volume of that particular acid that needs to be diluted to prepare a definite volume of dilute acid of the required concentration. Thus, the approximate concentration of such acids is given in the table below. 

 

Finding the volume of the concentrated acid required for preparation of dilute acid of various molarities:

It is possible to calculate the volume of this acid required for preparing a definite volume of the dilute solution if the molarity of the concentrated acid is known to us. The following molarity equation can be used to find out the molarity of the concentrated acid as described below:-

M₁V₁ = M₂V₂

Where, the molarity of the concentrated acid is M₁

V₁ is the volume of concentrated acid,

M₂ is the molarity of dilute acid,

V₂ is the volume of dilute acid.

For example, let us calculate the volume of concentrated sulphuric acid (molarity = 18M) required to prepare 250 ml of 2M acid.

According to the molarity equation:

M1V1 = M2V2

volume of concentrated acid required (V1) = ? molarity of dilute acid to be prepared (M2) = 2M

The volume of dilute acid to be prepared (V2) = 250 ml.

Substituting these values in the molarity equation, we get

18 x V1 = 2 x 250

V1=2 × 25018 = 27.8ml

Therefore, 27.8 ml of 18M H2SO4 must be diluted with water to make the volume 250 ml. The resulting solution will be 2M H2SO4.

Now, the molarity of concentrated acid (M1) = 18 M

Finding the volume of the concentrated acid required for preparation of dilute acid of various normalities:

It is possible to calculate the volume of this acid required for preparing a definite volume of the dilute solution if the normality of the concentrated acid is known to us. The following normality equation can be used to find out the molarity of the concentrated acid as described below:-

N1V1 = N2V2

where N1 is the normality of concentrated acid,

V1 is the volume of concentrated acid,

N2 is the normality of dilute acid,

V2 is the volume of dilute acid.

For example, let us calculate the volume of concentrated hydrochloric acid (normality = 12 N) required to prepare 250 ml of 4N acid.

According to the normality equation :

N1V1 = N2V2

Now, normality of concentrated acid (N1) = 12 N

The volume of concentrated acid required (V1) =?

normality of dilute acid to be prepared = N2 = 4N

The volume of dilute acid to be prepared (V2) = 250 ml.

Substituting these values in the normality equation

12 x V1 = 4 x 250

V1=4 × 25012 = 83.3ml

Therefore 83.3 ml of concentrated hydrochloric acid must be diluted with water to make 250 ml of 4 N HCl.

Finding the Molarity of Oxalic Acid in a 250 ml Solution

If the solution is already prepared, then a few ml of this solution has to be titrated with known strength of base (NaOH) solution using an indicator or with KMnO4. Then, use the formula V1S1 = V2S2 to calculate the strength.

 

Where V1 and V2 are the volumes of acid and base respectively, and S1 and S2 are the strength in the normality of acid and base, respectively. Now, divide the normality by 2 to get the molarity of oxalic acid. If the solution is about to be prepared, then find the mole number of oxalic acid.

 

The molecular weight of hydrated oxalic acid (COOH)2.2H2O is 126.

 

So, mole number – m = weight taken ÷ 126.

 

Since the volume is 250 ml, 1/4 of 1000 ml, the molarity = m × 4.

 

Equivalent Weight

Equivalent weight is used in the ratio and index calculations for various 2-component systems, such as epoxies and polyurethanes. Classically, it is defined as the molecular weight divided by functionality; that gives you a weight unit per reactive site essentially.

 

Suppose, you have a Side A having an equivalent weight of 135, and a Side B with an equivalent weight of 150. This means 135g of component A has the same number of reactive sites as 150g of component B and the ratio falls at 135:150. The ratio A: B, gives us a perfect stoichiometric balance (sometimes desirable and sometimes not).

 

In the case of a polyol, the equivalent weight is 56,100 divided by the ‘OH’ value. But, for isocyanates, it is 4,200, which is divided by the NCO weight percent.

 

Here, the epoxies are similar. For resin, it is the molecular weight divided by the number of epoxy groups. For the hardener/activator, it is the amine’s molecular weight divided by the number of active hydrogen atoms.