**HOW TO ** STANDARDIZE SOLUTION IN CHEMISTRY PRACTICALS

Before getting to the intended concept let's get a clear meaning of standardisation

Standardization is the process of determining unknown concentration of a solution by using the solution of known concentration.

The solution which is being standardized is called the titrant or analyte.

Standardization is done by just adding distilled water to the solution

Standardization is required if an intended(actual) titre value is not achieved after preparation of solutions. Before supplying solutions to students for practical activities, laboratory technician or teacher is required to assure if the solutions prepared give the exact or expected value of the experiment.

Sometimes happens while testing prepared solutions, you find the value obtained is either above or lower than the expected value, here is where standardisation applies.

HOW TO STANDARDIZE THE SOLUTION IF THE TITRE VALUE OBTAINED IS LESS THAN EXPECTED VALUE

For Example the titre value of experiment is 10cm3 but when performing experiment the obtained value is either less or more than 10cm3, to solve this scenario you have to perform standardization.

Let assume you were required to prepare 0.01M sulphuric acid to make 1500ml against 0.01M sodium hydroxide with a pipette of 20cm3 ,the actual value after titration was 10cm3, let say getting 8cm3 instead of 10 cm3 after titration, this signifies that the acid( 0.01M sulphuric acid) is more concentrated than its normal assuming the base is well prepared, therefore some amount of water should be added to dilute the acid.

To standardize a solution which is more concentrated is possible since you will be required to find an amount of water to add to the solution to return to its normal.

Follow the steps below to get it done

FIRST METHOD

I. Find a new molarity of sulphuric acid by using the formula below

From our assumptions

Molarity of acid(Ma)=? (the required new molarity)

Molarity of base(Mb) = 0.01M

Volume of acid(Va) = 8cm3 (the obtained value which is below than expected)

Volume of base(Vb)= 20 cm3

MaVa/MbVb=na/nb

Ma x 8/ 0.01 x 20= 1/2

Molarity of acid (Ma) is 0.0125M

The new molarity of sulphuric acid obtained from calculation (0.0125M) is absolutely more concentrated than 0.01M.

II. Find the volume of concentrated acid which should be mixed with water to make 1500 ml of 0.01M of the standardized sulphuric acid

From dilution law

McVc=MdVd

Mc= 0.0125M (the new molarity which is more concentrated than actual molarity)

Vc= ? ( volume of concentrated solution)

Md= 0.01M

Vd= 1500 cm3

0.0125 x Vc = 0.01 x 1500

Vc= 1200cm3

1200cm3 is the amount of 0.0125M sulphuric acid which is more concentrated than 1500 ml of 0.01M of the standardised sulphuric acid, therefore to get amount of water which should be added to dilute the acid you are required to subtract the concentrated volume(volume obtained from calculation which is 1200ml) from diluted volume which is 1500 ml to return the acid to normal.

1500ml - 1200ml

=300ml

300ml is the amount of water which should be mixed with 1200ml 0f 0.0125M sulphuric acid to make 1500ml of the 0.01M sulphuric acid.

SECOND METHOD

I. Find the volume of concentrated solution by taking the obtained value over the actual(expected) value times the volume of dilution.

8/10 x 1500ml

=1200ml (the amount of sulphuric acid needed to be standardised ).

II. Find the amount of water to be added by subtracting the obtained value from the actual value then multiplying with results with the volume of dilution.

10ml - 8ml

= 2ml

2/10 x 1500ml

= 300 ml( the amount of water to be added to make 1500 ml of 0.01M of standard sulphuric acid)

HOW TO STANDARDISE THE SOLUTION IF THE TITRE VALUE OBTAINED IS MORE THAN EXPECTED VALUE

let say getting 13cm3 instead of 10 cm3 after titration, this means that the base ( 0.01M sodium hydroxide) is more concentrated than its normal assuming the acid is well prepared, therefore some amount of water should be added to dilute base.

Follow the steps below to get it done

FIRST METHOD

I. Find a new molarity of sodium hydroxide by using the formula below

From our assumptions

Molarity of acid(Ma)=0.01m

Molarity of base(Mb) = ?(the required new molarity)

Volume of acid(Va) = 13cm3 (the obtained value which is more than expected)

Volume of base(Vb)= 20 cm3

MaVaMbVb=nanb

0.01 x 13 Mb x 20= 12

Molarity of base (Mb) is 0.013M

The new molarity of sodium hydroxide obtained from calculation (0.013) is absolutely more concentrated than 0.01M.

II. Find the volume of concentrated acid which should be mixed with water to make 1500 ml of 0.01M of the standardised sulphuric acid

From dilution law

McVc=MdVd

Mc= 0.013M (the new molarity which is more concentrated than actual molarity)

Vc= ? ( volume of concentrated solution)

Md= 0.01M

Vd= 1500 cm3

0.013 x Vc = 0.01 x 1500

Vc= 1153.9cm3

1153.9cm3 is the amount of 0.013M sodium hydroxide which is more concentrated than 1500 ml of 0.01M of the standardised sodium hydroxide, therefore to get amount of water which should be added to dilute the acid you are required to subtract the concentrated volume(volume obtained from calculation which is 1153.9ml) from diluted volume which is 1500 ml to return the base to normal.

1500ml - 1153.9ml

=346.1ml

346.1ml is the amount of water which should be mixed with 1153.9ml 0f 0.013M sodium hydroxide to make 1500ml of the 0.01M standardised sodium hydroxide.

SECOND METHOD

I. Find the volume of concentrated solution by taking the obtained value over the actual(expected) value times the volume of dilution.

10/13 x 1500ml

**=1153.9ml (the amount of sodium hydroxide needed to be standardised ).**

II. Find the amount of water to be added by subtracting the obtained value from the actual value then multiplying with results with the volume of dilution.

13 ml - 10ml

= 3ml

3/13 x 1500ml

**= 346.1 ml( the amount of water to be added to make 1500 ml of 0.01M of standard sodium hydroxide)**

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