Question

0. 99 g of acid was dissolved in water and the solution made up to 200 cm³, 20 cm³ of this solution required 15 cm³ of 0.105 N sodium hydroxide solution for complete neutralization. Find the equivalent mass of the acid.

Answer 1

Substituting we have 20 × N₁ = 15 × 0.105 

∴ N₁ = (15 x 0.105)/20 = 0.07875

∴ Normality of acid solution = 0.07875 

Mass of the acid in one dm³ of the solution – 0.99 × 5 g = 4.95 g 

Eq. mass of the acid is given by E = W/N = 4.95/0.07875 = 62.85