0.968 g of an acid are present in 300 ml of a solution. 10 ml of this solution. required exactly 20 ml of 0.05 N KOH solution. calculate no. of neutralisable protons and equivalent weight of the acid. (given molecular wt. of acid is 98)

Consider the number of neutralisable protons to be n.
     Molecular weight of the acid = 98
So,
     Equivalent weight = 98 / n
Thus,
     Number of equivalents of the acid = (0.968) / (98/n)
                                                             = 0.00988 n
Therefore,
     Number of equivalents of acid in 10 mL = 0.00988 n / 30
Thus, we have,
              0.00988 n / 30
= 0.05 * 20 / 1000
Solving, we get
     n = 3
Hence,
     Number of neutralisable photons = 3
     Equivalent weight of the acid       = 98 / 3 = 32.67 g

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