Preparation of standard, Normal and molar solutions Subscribe to my youtube channel for such great contents www.youtube.com/c/AwesomeBiochemistry

1. Preparation of Standard,
normal and molar solutions
Dr. Kalpesh

2. Percent Solutions
 w/w  10% w/w solution contain 10 gm of solute into 10 gm of
solvent
w/v  10% of w/v solution contains 10 gm of solute in 100 ml of
solution (not the solvent)
v/v  10% v/v solution contains 10 ml of concentrate per 100 ml
of solution (not the solvent)

3. Avogadro's Number  6.022 x 1023 per Mole.
• Avogadro’s Number is number of atoms (or molecules) in 1
mole of substance.
• Problem #1: 0.450 mole of Fe contains how many atoms?
• 0.450 mol x 6.022 x 1023 mol¯1
• Problem #2: 0.200 mole of H2O contains how many molecules?
• 0.200 mol x 6.022 x 1023 mol¯1

5. Use of Molecular Weight
Molecular weight is important, because it connects
The Macroscopic scale,
Where we all live
The Microscopic scale,
Where chemistry occurs
Gms of a substance Number of molecules in
mole
Use the molecular weight to connect between the two
scales

6. • Calculate the number of moles in 1.058 gram of H2O. (MW of H2O=18
gm/mole)
1.058𝑔𝑚
18𝑔𝑚/𝑚𝑜𝑙𝑒
• Calculate the number of molecules in 1.058 gram of H2O
1.058𝑔𝑚
18𝑔𝑚/𝑚𝑜𝑙𝑒
𝑥 Avogadro′
s number

8. Determination of molecular weight (H2SO4)
• Make a list of each element and the number of atoms of each element
present in the substance (2 H, 1 S, 4 O)
• Go to periodic table and determine the atomic weight of each element.
H 1.00794
S 32.066
O 15.9994
• Multiply each atomic mass by the number of atoms in the formula.
H 1.00794 * 2 = 2.015
S 32.066 * 1 = 32.066
O 15.9994 * 4 = 63.998
• Add up the results of above step
2.015+32.066+63.998 = 98.079 = molar mass of sulfuric acid

10. Molarity
• Number of moles of solute in 1 litre of solution.
• Molecular weight = gms/mole
1. Prepare 2M solution of NaCl (mw = 58)
• 1 mole = 58 gm  2 mole = 116 gms of Nacl
• 116 gms of NaCl in 1litre solution
2. Prepare 100 ml of 2M solution of NaCl
•
𝑤1
𝑣1
=
𝑤2
𝑣2
→
116𝑔𝑚
1000𝑚𝑙
=
𝑤2
100𝑚𝑙
→ 𝑤2 = 11.6 𝑔𝑚
• 11.6 gm of NaCl in 100 ml of solution
MW (gms/mole)
Na = 23
Cl = 35.5

11. Normality
• It is the number of gram equivalent of solute dissolved in 1 litre of
solution
• It is based on chemical reaction
• E.g. 1M solution of H2SO4  1 mole of H2SO4 in 1 litre of solution
But it gives 2 moles of acid
So, 1M of H2SO4 will be equivalent to
2 moles of acid

12. Normality (Cont…)
• Normality represents the molar concentration ‘only of the Acid
Component (H+ for Acid)’ or ‘only the base component (OH- for
base)’
• Finally,
• N = M x Number of H+ or OH- ions
• 2M H2SO4 = 4N
• 2M H3PO4 = 6N
• 2M HCl = 2N
• 2M NaOH = 2N

13. Ideal procedure of calculating Normality
1. Calculate compound’s equivalent mass
• This is done by taking compound’s MW and dividing by the
number of H+ ions or OH- ions
• NaOH’s equivivalent mass = 40/1
• H2SO4’s equivivalent mass = 98/2
2. Apply formula
• Grams of compound needed = (Desired N)x(Equivalent mass)x(Volume in Litres desired)
• Prepare 250 ml of 2N NaOH
• 2N * 40 * 0.250 = 20 gm

14. If compound is liquid then follow below
formula
• 𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑐𝑜𝑛𝑐𝑒𝑛𝑡𝑟𝑎𝑡𝑒𝑑 𝑎𝑐𝑖𝑑 𝑛𝑒𝑒𝑑𝑒𝑑 =
𝑔𝑚𝑠 𝑜𝑓 𝑎𝑐𝑖𝑑 𝑛𝑒𝑒𝑑𝑒𝑑
% 𝐶𝑜𝑛𝑐𝑒𝑛𝑡𝑟𝑎𝑡𝑖𝑜𝑛 𝑥 𝑆𝑝𝑒𝑐𝑖𝑓𝑖𝑐 𝑔𝑟𝑎𝑣𝑖𝑡𝑦
• Prepare 200 ml of 0.2 N H2SO4 solution
1. MW = 98, number of H+=2 so eq.wt=49
2. Gms of Compound needed = 0.2 x 49 x 0.2 = 1.96 gm
3. Volume of concentrated acid needed =
1.96
0.98𝑥1.84
= 1.09 𝑚𝑙

15. Make 1 Litre of 1M aqueous solution of H2SO4
• Read the label on the bottle of H2SO4 reagent. (18M)
•
1𝑚𝑙
0.018𝑚𝑜𝑙𝑒𝑠
=
𝑥𝑚𝑙
1𝑚𝑜𝑙𝑒
→ 𝑥 = 55.6 𝑚𝑙 of H2SO4 contains 1M of it
• So, We slowly add 55.6 ml of the H2SO4 to about 500 ml of DI
water and then top it off with more water to make it 1 Litre
• Caution – Never add water into a large volume of concentrated
acid… you risk creating explosion
• Remember – water into acid  Blast

16. Typical concentrations of concentrated acid
& Bases (As on Lable)
Name Wt% Density (Sp. Gr)
Gm/ml
MW
Gm/Mole
Molarity
Acetic acid 99.7 1.05 60.05 17.4
Ammonium Hydroxide 35
(Aqueous ammonia) 28 0.89 17 14.6
Hydrochloric acid 37 1.18 36.46 12
Nitric acid 70 1.40 63 15.6
Phosphoric acid 85 1.69 14.7
Sulfuric acid 96 1.84 98.07 18
Density = weight of 1 ml liquid
𝑀𝑜𝑙𝑎𝑟𝑖𝑡𝑦 =
𝐷𝑒𝑛𝑠𝑖𝑡𝑦 𝑥 𝑊𝑡% 𝑥 10
𝑀𝑊

17. Shortcuts
• 𝑀𝑜𝑙𝑎𝑟𝑖𝑡𝑦 𝑁𝑜𝑟𝑚𝑎𝑙𝑖𝑡𝑦 =
𝑋𝑔𝑚 𝑥 𝑛
𝑀𝑊 𝑥 𝑉𝑙𝑖𝑡
• For Molarity, n=1
• For Normality, n=calculated
• 𝑚𝑙 =
𝑋𝑔𝑚
𝑆𝑝.𝐺𝑟. 𝑥 %

18. Molar solutions from powder
• How much sodium chloride is needed to make 1 litre of an
aqueous 1M solution (MW=58.5 gms/mole)
• How much sodium chloride is needed to make 1 litre of an
aqueous 2M solution (MW=58.5)
• How much sodium chloride is needed to make 1 litre of an
aqueous 0.1M solution

19. Molar solutions from liquid
• Make 1 litre of 1M aqueous solution of H2SO4 (MW=98.07, Sp. Gr. =
1.84, Purity=96%)

20. Normal solutions from powder
• Calculate the normality of a NaCl solution prepared by dissolving
2.9216 gms of NaCl in water and then topping it off with more
water to a total volume of 500 ml (MW=58.44)

21. Normal solution from liquid
• Calculate the volume of concentrated aqueous sulphuric acid
having Sp.Gr. 1.842 & containing 96% H2SO4, required to prepare 2
litre of 0.20N H2SO4
• Calculate the volume of concentrated HCl, Having a density of
1.188 gm/ml and containing 38% HCl by weight, needed to prepare
2 litre of 0.20N HCl solution (MW=36.461)

22. Thanking You