Question

what will be the volume of 0.1M Ca(OH)2 required to neutralize 10 ml of 0.1 N HCl

Answer 1

n-factor of Ca(OH)₂ = 2 (because when dissociated, it gives out 2 OH⁻ ions)

Normality = Molarity x n-factor = 0.1 x 2 = 0.2 = N₁

Volume of Ca(OH)₂ required = V₁

Normality of HCl = 0.1 = N₂

Volume of HCl = 10ml = V₂

Using the formula N₁V₁ = N₂V₂

(0.2)(V₁) = (0.1)(10)

V₂ = 1/0.2

V₂ = 5ml

Thus, 5ml of 0.1M Ca(OH)₂ is required to neutralise 10ml of 0.1N HCl.

Hope it helps!!