1.26 g of a dibasic acid were dissolved in water and the solution made up to 200 mL. 20 mL of this solution were completely neutralised by 10 mL of `N/5` NaOH solution. Calculate the equivalent mass and molecular mass of the acid.

Explanation

0.30 g of the given acid are present in 100 mL solution.
` therefore ` Mass of acid present in 15 ml `= (0.30 )/(100) xx 5 = 0.045 `g
This mass of the acid neutralises 14.3 mL of `(N )/(20 ) `NaOH solution.
Suppose, the equivalent mass of the acid is E.
Number of gram equivalents of acid `= (0.045 )/(E )`
Number of gram equivalents of NaOH `=(1/20 xx 14.3 )/(1000)`
The number of gram equivalents of acid must be equal to the number of gram equivalents of base. Therefore,
`(0.045 )/(E ) = ( 1/20 xx 14.3 )/(1000)`
or ` E= (0.045 xx 1000 xx 20 )/(14.3 )= 62.94 `
Molecular mass of the given acid `=E xx ` basicity
` = 62.94 xx 2 =125.9`