Chemist Interview Question:

What are the method for the preparation of 1 normal solution of hydrochloric acid?

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Density of HCl: 1.48 g/ml
Molecular weight of HCl: 36.5g/mole
Concentration of HCl stock: 36%
Concentration of another HCl stock: 12N
Final volume: 1L

Let's assume that we're making a 1N HCl solution. Using the 36% concentration stock, we'll need to find out how many ml we would need. However, HCl is interesting because it has a higher density than water.

For a 1N solution (also known as 1M) however, you would need this amount of grams:

(1M HCl) x (36.5 g/ mole HCl) x (1L) = 36.5 grams.

Now we need the amount of ml from a 36% solution. This is different from having a known concentration like 10M, for example.

36% = 36 grams HCl / 100 grams of stock concentration solution

(36.5 grams HCl) x (100gram of stock solution/ 36 grams HCl) x (1 ml/ 1.48 grams) =
68.5ml

Dissolve 68.5ml of 36% solution into 931.5ml of ddH20.


If your stock concentration came in 12M, then using the following equation:
M = concentrations and V = volume.

M1V1 = M2V2
V1 = (M2V2 )/M1


Using the second equation,
V1 = volume of your stock concentration
M1 = concentration of your stock
V2 = volume of your final preparation
M2 = concentration of your final

V1 = (1 M x 1L )/(12M) = 0.083 L = 83ml

Dissolve 83 mL of 12M HCl into 917ml of ddH2O
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