Question

How would you make up 2.5L of 5N KOH? (VALENCE 1)

How would you make up 50mL of 3.5 N H3PO4?

How would you make up 7000mL of 1N HCL?

Answer 1

In these following problems, the basic concepts of Solution Chemistry, normality and gram-equivalent weight of the compounds are used to solve the problems.

1) In the first problem, we have to prepare KOH solution. The definition of Normal solution states that 1 gm-equivalent of compound(solute) has to be dissolved in 1 litre of solvent. Here, we are taking KOH as the solute. The mol-equivalent weight of KOH is 56 g/mol (reference: Wikipedia). The donatable OH^- present in KOH is one. So, the gram-equivalent weight of a base or acid = (mol-equivalent weight / no of donatable H⁺ or OH^- present in that base or acid).

As the donatable OH^- in KOH is only one, the gram-equivalent and mol-equivalent weight of KOH are same. So, we need 56 gram of KOH in 1 litre of solvent to prepare 1 (N) solution. But, we need 2.5 L of 5 (N) solution. So, we need (56*2.5*5) = 700 gram of KOH to prepare the solution.

2) In the 2nd problem, the total number of donatable H⁺ in H₃PO₄ is three. So, the gram-equivalent weight of H₃PO₄ = (mol-equivalent weight / 3). The mol-equivalent weight of H₃PO₄ is 98 g/mol (reference: Wikipedia). So, the gram-equivalent weight of H₃PO₄is = (98/3) = 32.667 gram. This amount of H₃PO₄is needed in 1 litre of solvent to prepare 1 litre 1(N) solution of H₃PO₄. Here, we need 50 mL 3.5 N H₃PO₄solution.

1 litre = 1000 mL solvent needs 32.667 g of H₃PO₄to prepare 1(N) solution

So, 50 mL solvent needs (50/1000)*32.667 g = 1.633 g of H₃PO₄to prepare 1(N) solution

So, 50 mL solvent needs (1.633*3.5) g = 5.717 g of H₃PO₄to prepare 3.5(N) solution.

3) In the 3rd problem, we have to prepare 7000 mL of 1(N) HCl. As, the donatable H⁺ in HCl is only one, the gram-equivalent weight and mol-equivalent weight are same like KOH. The mol-equivalent weight of HCl is 36.5 g/mol (reference: Wikipedia), which is also the gram-equivalent weight of HCl.

1 litre of solvent needs 36.5 g of HCl to prepare 1(N) solution

So, 7000 mL = 7 litre of solvent needs (36.5*7) = 255.5 gram of HCl to prepare 1(N) solution.

Simple cross-multiplication is used in the calculation part to solve the problems.