0.1914 g of an organic acid is dissolved in 20 ml of water,25 ml of 0.12 N NaOH is required for complete neutralisation of the acid solution. The equivalent weight of the acid is

Asked by Anil | 16th May, 2017, 10:16: PM

Expert Answer:

Moles of acid = 0.1914/M

Moles of NaOH = 25 x 10-3 x 0.12 = 3 x 10-3

Now, mole of acid = mole of NaOH

Or 0.1914/M = 3 x 10-3 or M = 0.1914/3x 10-3

                          = 63.8 gmol-1

Answered by Vaibhav Chavan | 17th May, 2017, 11:44: AM