Titration lab report. Aim: to determine the quantity and concentration of NaOH to completely neutralize one molar hydrochloric acid till the n point.

Method: arrange burette, stand and clamp in the following way and place the conical flask

• Using a pipette and a pipette pump take 20 ml of HCl in the conical flask
• Add few drops of phenoptheline to acid
• Add NaOH in the burette till the 0 level
• Slowly add NaOH to HCl
• Stir the solution after adding every drop till the solution changes the colour

Observations:

Remember:

• were lab coats during the experiment

• conduct all the experiment at same time and place to keep the temperature and light intensity constant
• clean all the apparatus with distilled water before every experiment

Calculation:

Uncertainties:

Burette ± 0.5

Pipette ± 0.05

Volume ± [(0.5/10) + (0.05/10)]x10 cm-3

± [0.05+0.005]x10

± 0.6

Concentration ± 0.1 moles dm-3

No of moles ± [(0.1/0.6)+(0.6/8.2)]x0.005

± 0.001 moles

Concentration= no of moles / volume in dm3 

0.5= X/ 0.010

X= 0.5x0.010

X=0.005 moles

Mole ratio= 1:1

Therefore 0.005 moles of NaOH is required to neutralize 0.005 moles of HCl

Thus concentration for the exp with 10 cm3= 0.005/0.0082

=0.6 ± 0.1 moles dm-3 

Therefore concentration for exp with 15 cm-3 is: 0.6 ± 0.1 moles dm-3

And concentration for exp with 20 cm-3 is: 0.6 ± 0.1 moles dm-3

Conclusion: The colour of the indicator turns pink in basic solution and the drop before the colour change is supposed to be the n point. This means that it is the point at which the solution is neutral. Using which the number of moles can be calculated so can the concentration which came to the results of no of moles as 0.005 moles in first experiment and the concentration which is constant to all the experiment comes to 0.6 ± 0.1 moles dm -3