Question

The titration of 0.1495 g of dry potassium acid phthalate (mm=204.227 g/mol) requires 42.59 ml of naoh solution. what is the concentration of the naoh solution?

Answer 1

M1V1 =M2V2

Where M1 and M2 is the molarity of the first and second solution respectively.

And V1 and V2 is teh volume of the first and tehs econd solution respecyively.

Molarity = No of moles/ 1 L of solution

No of moles = mass of the substance/molar mass of the substance

No of moles of potassium acid phthalate = 0.1495/204.227 = 0.00073 moles

Thus the same number of moles of NaOH is required in order to reach the equivalence point.

As Molarity = No of moles/ 1 L of solution

Molarity of NaOH solution = (0.00073/42.59) x 1000= 0.017 M