Dry potassium acid phthalate (KHC8H404) for one hour at 110°C and cool. Accurately weigh 2.00 to 2.05 gms of the sample to four significant figures and transfers to a 100ml volumetric flask. Dissolve the salt using distilled water and make the solution up to the mark. Pipette out 20ml of the above solution in an Erlenmeyer flask. Add 2 drops of phenolphthalein in a indicator and titrate with the NaOH solution taken in the burette. The end point is the appearance of a faint permanent pink colour. Repeat the experiment to get concordant values. From the volume of the NaOH solution, calculate its molarity. Using this standard NaOH solution titrate 20ml potions of the given HCI solution using phenolphthalein indicator. Calculate the molarity of the CHI solution. Weight of potassium acid phthalate # Of moles of potassium acid phthalate Molarity of KHC8H4H4O4 solution = n/01 = . # of moles of KH in 20ml = 0.02 x molarity ...... X moles of KHC8H4O4 solution neutralizes x moles of NaOH solution. Let the overage volume of NaCH solution added be Y ml. gms (w) (n) M (X) Hence the molarity of NaOH = 1000x Molarity. Y A similar calculation can be performed for the calculation of the molarity of HCL.