Question

25 ml of kOH solution was neutralized by 30ml of HCL of normality 0.01.Find the normality of KOH

Answer 1

The reaction taking place is:

KOH + HCl ------------> KCl + H2O

So, 1 part of KOH reacts with 1 part of HCl.

=> As 30 mL of HCl of normality 0.01 N was required.

Basicity of HCl (no. of H+ ions formed per molecule on dissociation) = 1

Normality = Basicity x Molarity

0.01 = 1 x Molarity

Molarity of HCl solution = 0.01 M.

=> 30 mL = 0.03 L

Molarity = (moles of HCl)/(litres of solution)

0.01 = moles/0.03

Moles of HCl =  0.0003 moles of HCl

So, 1 part is equal to 0.0003 moles (from equation).

=>So, 0.0003 moles of KOH is required.

25 mL or 0.025 L of KOH solution has 0.0003 moles of KOH in it.

Molarity = 0.0003/0.025 => 0.012  M KOH solution.

Acidity of KOH = 1 ( 1 OH-1 ion per molecule of KOH in aqueous state)

Normality = Molarity x Acidity

Normality = 0.012 x 1 => 0.012 N KOH solution

Hence, the normality of KOH solution is 0.012 N.