How to Dilute High Concentration Acid

For Nitric acid (HNO3)

On Nitric acid bottle label, there’re information on
i) specific gravity =>> 1 liter = 1.4 kg
ii) M (jisim molekul relative) =>> 63 g/mol
iii) Concentration =>> 65%


CALCULATION

From specific gravity we know;
1 Liter HNO3 = 1400 g HNO3 for 100% concentration

Thus, mass of 65% HNO3 (as stated at bottle label)
= % x m
= 65 / 100 x 1400g (obtain from specific gravity)
= 910 g

Mole no. for 65% nitric acid in 1 liter
= m / M
= (910 g) / (63 g/mole)
= 14.44 mole

Thus, molarity of 65% nitric acid in 1 liter is 14.44 M

Now, we know that 65% nitric acid is 14.44 M (Mn), but the question is “how its molarity for 2% nitric acid?” (Mm). To find out this answer, we can calculate the molarity for 2% nitric acid is as below;

Let say the molarity for 2% nitric acid is Mm, thus We can obtain Mm by calculation as below

Mm / 14.44 = 2 / 65
Thus Mm = 0.44, so molarity for 2% nitric acid is 0.44 M (Mm)

For dilution purpose, let say we need 100 ml of nitric acid solution (Vm). Thus the volume of nitric acid from the bottle (Vn) that needs to dilute from 65% to 2% is as follow;

We can use the information as calculated before; Mn = 14.44 M; Vn = ??? ; Mm = 0.44 M; Vm = 100 ml

MmVm= MnVn
Vn = (0.44 M).(100 ml) / (14.44 M)
= 3.05 ml
Thus the volume of nitric acid from the bottle (Vn) that needs to dilute from 65% to 2% is 3.05 ml #