This document describes a chemistry lab experiment to determine the concentration of HCl using conductometric titration with NaOH. The key steps are: 1) 40ml of HCl is titrated with 0.1M NaOH solution while measuring the conductance. 2) The conductance is plotted against the volume of NaOH added. 3) The neutralization point is determined from the graph by extrapolating the straight lines. 4) Using the titration data and volume of HCl, the molarity of HCl is calculated to be 0.015M.

1. NAME: U.SHALEM RAJU
ROLL .NO :-227R1A1256
CHEMISTRY LAB
ESTIMATION OF HCL USING
CONDUCTROMETRIC TITRATION

2. AIM:-
•To determine the strength of HCl by using
standard NaOH. Using conductivity metre.
•REQUIREMENTS:-
•Conductometer , conductivity cell ,
distilled water , 250ml beaker , stirrer ,
HCl , 0.1m NaOH .

3. PRINCIPLE:-
As the alkali is added from the burette into the cell
Containing acid , the concentration H^+ ions change
In a graphical manner , which leads to a considerable
change in the electrical conductance of the solution .
Which is measured using a conductivity meter .
HCl + NaOH NaCl + H2O

4. • Then from the plot of conductance versus
volume of alkali , the precise neutralization
point is determined. the titration of strong acid
verses strong acid versus strong base involves
the following equation. In the titration of HCl
against NaOH , initially the conductance of HCl
solution is very high due to complete
ionization.

5. •As the alkali is added , the conductance of
solution decreasing and after the
neutralization point the conductance starts
increasing.
• This is because of the addition of the
alkali , the fust moving H^+ ions are
replaced by Na^+ ions (slow moving) once
the neutralization point is reached addition
of alkali introduces fast moving OH^- ions

6. PROCEDURE:-
•Take 40ml of HCl into a clean 100ml
beaker . Fill the burette with 0.1m NaOH
solution dip a conductivity cell in the HCl
solution and connect it to the
conductivity meter and note the initial
conductance of solution . Rundown the
NaOH solution from the conductance

7. OBSERVATION:-
S.NO Volume of NaOH (ml) conductance
1. 0 2.0
2. 1 1.8
3. 2 1.6
4. 3 1.4
5. 4 1.2
6. 5 1
7. 6 0.8
8. 7 1.0
9. 8 1.2
10. 9 1.4
11. 10 1.6
12. 11 1.8

8. • A plot of conductance on Y- axis versus volume of
alkali on X-axis is drawn. Extrapolate the straight lines
and note the neutralization point of the graph.

9. CALCULATION:-
M1V1 = M2V2
• M1 = molarity of NaOH solution = 1M
• V1 = volume of NaOH solution rundown=
burette reading (end point)
• V2 = volume of HCl = 40ml.
• M2 = molarity of HCl solution = 6
• M2 = m1v1/v2 = 0.1×6/40 = 0.015M
• Amount of HCl = m2 × 36.5
• 0.015 × 36.5 = 0.54

10. THANK
YOU!