Question

CONDUCTOMETRIC TITRATION OF STRONG ACID Vs STRONG BASE

AIM

To determine the amount of hydrochloric acid present in the whole of the given solution

by conductometric titration using standard sodium hydroxide solution of 0.2 N.

PRINCIPLE

Solutions of electrolytes conduct electricity due to the presence of ions. The specific

conductance of a solution is proportional to the concentration of ions in it and the mobility of

the ion. The reaction between HCl and NaOH may be represented as,

HCl + NaOH NaCl + H2O

When a solution of hydrochloric acid is titrated with sodium hydroxide, the fast

moving hydrogen ions are progressively replaced by slow moving sodium ions. As a result

conductance of the solution decreases. The conductance decreases until the end point is

reached. Further addition of alkali raises the conductance sharply as there is excess of

hydroxide ions. A graph is drawn between volume of NaOH added and the conductance of

solution. The exact end point is the point of intersection of the two lines.

PROCEDURE

The burette is filled with sodium hydroxide solution up to the zero level. 20ml of the

given HCl is pipetted out into a clean 100ml beaker and then diluted to 50ml by adding

conductivity water. The conductivity cell is placed in it. The electrodes are well immersed in

the solution. The two terminals of the cell are connected with a conductivity meter.

Now 1ml of NaOH from the burette is added to the solution taken in the beaker,

stirred for some time and then conductivity is measured. (The conductivity is going on

decreasing up to the end point). This process is repeated until five readings are taken after the

end point has been reached. Now the graph is plotted by taking volume of NaOH in the Xaxis and conductance in the Y-axis. The end point is noted from the graph and the amount of

hydrochloric acid present in the whole of the given solution is calculated.

RESULT

The amount of HCl present in the whole of the given solution = ------- g

Calculation

Volume of HCl V1 = 20 ml

Strength of HCl N1 = ?

Volume of NaOH V2 (obtained from graph) = …………. ml

Strength of NaOH N2 = 0.2 N

According to the law of volumetric analysis,

V1N1=V2N2

N1 = V2N2 / V1



Strength of HCl = …….. N

Calculation of amount of HCl

The amount of HCl present in 1000 ml of the given solution=Equivalent weight x strength



= 36.46 x ………



= …… g ​

Answer 1

Answer:

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Answer 2

Answer:

Let us consider the titration of strong acid, like HCI, with a strong base like NaOH.

Before alkali is added, the acid solution has a high conductance due to the highly mobile hydrogen ions.

When alkali is added to a strong acid, the highly mobile hydrogen inns are replaced by less mobile metallic ions: the conductance will, therefore, decrease. When neutralization is complete, further addition of alkali will cause the conductance to increase owing to the excess of cation and highly mobile OH^-ions; the conductance will thus be minimum at the equivalence point. A plot of conductance versus the volume of titrant added will consist of two almost straight line branches intersecting at the neutralization point. When both acid and base are strong, the titration can be carried out from either side, ir., alkali can be added to the acid or vice-versa.

Calulation: -

Volume of Hcl(V1)= 20 ml

Strength of HCl (N1) = ?

Volume of NaOH (V₁) = x mL

Strength of NaOH (N2) = 0·2N

According to law of Volumetric analysis

V1N1 = V2 N2

NI=V2N2/V1

N1=x * 0.2/20

N1= y N

Strength of HCl =y N

Calculation of amount of HCl

The amount of HCl present in 1000mL of given solution = equivalent weight *strength

= 34·96*y

=z g

For more references:-

https://brainly.in/question/14475540?

https://brainly.in/question/8464729?

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