Illustrative Examples

1. What volume of 0.05 M H2SO4 will be required to completely neutralize 15 ml of 0.2 N NaOH solution ? Solution H2SO4 is a dibasic acid. So Normality = 2 × Molarity. Normality of H2SO4 = 2 × 0.05 = 0.1 N N1V1 (H2SO4) = N2V2 (NaOH ) or 0.1 ×V1 = 0.2 × 15 ∴ V1 = 3 / 0.1 = 30 m30 ml of 0.05 M H2SO4 will be required.

2. 10 ml of HCl are required to neutralize completely 25 ml of 0.1 M Na2CO3 solution. What is the strength of HCl solution in grams per liter ? Solution Na2CO3 is a diacidic base. So Normality = 2 × Molarity. Normality of Na2CO3 = 2 × 0.1 = 0.2 N. N1V1 (HCl)= N2V2 (Na2CO3) or N1× 10 = 0.2 × 25 ∴ N1 = 5 / 10 = 0.5 N Normality × Equivalent weight = Strength in grams per liter. 0.5 × 36.5 = 18.25 . The strength of HCl solution is 18.25 grams per liter.

3. How many ml of 0.025 N HCl solution will be required to neutralize 25 ml solution prepared by dissolving 1.4 grams KOH in 500 ml solution ? Solution Strength of KOH solution = 1.4 grams in 500 ml. It means 2.8 grams per liter. Grams per liter = Normality × Equivalent weight. 2.8 = N1× 56 . N1= 2.8 / 56 = 0.05 N . N1V1( HCl )= N2V2(NaOH ) or 0.025 ×V1= 0.05 × 25 ∴ V 1 = 50 ml . 50 ml HCl solution will be required.

4. A solution of NaOH was prepared by dissolving one gram NaOH in 250 ml water solution. What volume of 0.025 N solution of HCl will be required to neutralize 25 ml NaOH solution ? Solution Strength of NaOH solution = 1 gram in 250 ml. It means 4 grams in 1 liter. Grams per liter = Normality × Equivalent weight 4 = N1× 40 or N1 = 4 / 40 = 0.1 N . N1V1(HCl)= N2V2(NaOH) or 0.025 ×V1 = 0.1 × 25 or V1 = 2.5 / 0.025 = 100 ml 100 ml HCl will be required.

5. How many grams of NaOH are equivalent to 1.25 ml of 0.1 N hydrochloric acid ? Solution The reaction between HCl and NaOH is a neutralization reaction. 1.25 ml of 0.1 N HCl will neutralize 1.25 ml of 0.1 N NaOH solution. 1000 ml 1 N solution of NaOH contains 40 grams NaOH. 1000 ml 0.1 N solution of NaOH contains 4 grams NaOH 1.25 ml 0.1 N solution of NaOH contains 5 x 10−3 grams NaOH 5 x 10−3 grams of NaOH are equivalent to 1.25 ml of 0.1 N HCl solution.

6. If 25.2 ml of 0.2703 N acid reacts with 35.1 ml of base , what is the normality of the base ? Solution N1V1(Acid ) = N2V2 (Base) 25.2 x 0.2703 = N2× 35. 1 N2 = 0.1941 The normality of the base 0.1941 N