STANDARD SOLUTIONS,primary standard, secondary standard

STANDARD SOLUTIONS, Primary standard, Secondary standard

STANDARD SOLUTIONS

Primary standard

Secondary standard 

 

STANDARD SOLUTIONS

There are two types of standard solutions in titrimetric (volumetric) analysis—the primary and the secondary standard solutions.

The primary standard 

solution is the reference for the volumetric analysis and therefore, it should be

carefully prepared. An ideal chemical for primary standard should satisfy the following conditions: 

(i) It must be stable, easy to obtain and to dry, and to preserve in pure form. 

(ii) It must have a large molecular mass to minimise the errors in weighing. 

(iii) It should not absorb moisture from the air during weighing.

(iv) It should be easily soluble in water. 

(v) It should not result in the production of interfering substances or reactions with the titration process. 

Note A perfect primary standard is difficult to prepare and so, any substance that meets most of the requirements should be used.

The secondary standard 

solution is usually prepared to approximate concentration and then titrated against a primary standard to get the exact concentration. This is because the substance used for preparing the secondary standard solution is not very stable and cannot be accurately weighed. Examples of secondary standard are strong acids such as hydrochloric acid; or strong bases such as sodium hydroxide.

I. To prepare 0.1N standard solution of sodium hydroxide.

A strong base such as sodium hydroxide (NaOH) should be standardised against a weak acid such as oxalic acid (COOH)2. 

(i) To prepare accurately 0.1N solution of oxalic acid as the primary standard: Weigh 6.303 g of oxalic acid accurately, dissolve it in distilled water, and make up the volume up to 1 litre with distilled water. 

(ii) To prepare approximately 0.1 N solution of NaOH: Dissolve approximately 4.0 g of NaOH in 1 litre of distilled water. 

(iii) To determine exact strength of the NaOH solution: Place 10 ml of 0.1 N oxalic acid solution in a 100 ml conical flask. Add 2 drops of phenolphthalein indicator (1 % solution in 50 % ethanol).

Run the NaOH solution gradually from a burette till a light pink end point is reached. 

Note the reading on the burette. Repeat the same procedure two more times and find the average. For example, Y ml of the NaOH solution neutralise 10 ml of 0.1 N oxalic acid. 

 

Therefore, from the equation NV = NzV2, the normality of

(iv) Dilution Factor (D.F.): If the normality of nNaOH is more than 0.1, the dilution factor should be determined as:

D.F=

Then, (1000 x D.F.) ml of NaOH + [1000 - (1000 X D.F.)] ml of distilled water = 0.1 N NaOH.

For example, let normality of NaOH obtained be equal to 0.11.

Therefore, 0.1 N NaOH = (1000 x 0.9) ml of NaOH +1000-(1000 x 0.9) ml of distilled water

i.e., 900 ml of 0.11N NaOH + 100 ml of distilled water will give 1000 ml of 0.IN NaOH.

II. Standardisation of 0.1 M hydrochloric acid (HCI) with anhydrous sodium carbonate (Na2CO3). 

(i) To prepare approximately 0.1M hydrochloric acid (HCI) the volume of acid required to make IM HCI can be calculated from the specific gravity and the percentage composition of the concentrated acid which are indicated on the label.

Sp. gr. of conc. HCl = 1.18, i.e., 1 ml HCI weighs 1.18 g.

Percentage composition (w/w) = 35.4 %, i.e., 100 ml of concentrated acid solution contains 35.4 g of acid.

Molecular weight of HCl = 36.5. Then, 36.5 g of acid will be present in 

Therefore, if 87.3 ml of conc. HCl is diluted to 1 litre, it will give 1 M HCI.

Thus, to make approx. 0.1 M HCl, dilute 9 ml of conc. HCl to 1 litre with distilled water.

Caution: 

Always remember to add acid to water, and not water to acid.. 

(ii) Standardise this HCl solution by titrating against the weak base sodium carbonate (Mol. wt. 106). Adjust the strength of the acid to 0.1M after calculating the dilution factor as in the first example for NaOH. 

Note In the same way, a molar solution of any acid can be prepared by using the equation.