Preparation of Standard, normal and molar

solutionsDr. Kalpesh

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)

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

Use of Molecular Weight

Molecular weight is important, because it connectsThe Macroscopic

scale,Where we all live

The Microscopic scale,Where chemistry occursGms of a substance Number of molecules in mole

Use the molecular weight to connect between the two scales

• Calculate the number of moles in 1.058 gram of H2O. (MW of H2O=18 gm/mole)

• Calculate the number of molecules in 1.058 gram of H2O

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 1.00794S 32.066O 15.9994

• Multiply each atomic mass by the number of atoms in the formula.H 1.00794 * 2 = 2.015S 32.066 * 1 = 32.066O 15.9994 * 4 = 63.998

• Add up the results of above step2.015+32.066+63.998 = 98.079 = molar mass of sulfuric acid

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

• 11.6 gm of NaCl in 100 ml of solution

MW (gms/mole)Na = 23Cl = 35.5

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

solutionBut it gives 2 moles of

acidSo, 1M of H2SO4 will be

equivalent to 2 moles of acid

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

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

If compound is liquid then follow below formula

𝑉𝑜𝑙𝑢𝑚𝑒𝑜𝑓 𝑐𝑜𝑛𝑐𝑒𝑛𝑡𝑟𝑎𝑡𝑒𝑑𝑎𝑐𝑖𝑑𝑛𝑒𝑒𝑑𝑒𝑑=𝑔𝑚𝑠𝑜𝑓 𝑎𝑐𝑖𝑑𝑛𝑒𝑒𝑑𝑒𝑑

% 𝐶𝑜𝑛𝑐𝑒𝑛𝑡𝑟𝑎𝑡𝑖𝑜𝑛𝑥𝑆𝑝𝑒𝑐𝑖𝑓𝑖𝑐 𝑔𝑟𝑎𝑣𝑖𝑡𝑦

• Prepare 200 ml of 0.2 N H2SO4 solution1. MW = 98, number of H+=2 so eq.wt=492. Gms of Compound needed = 0.2 x 49 x 0.2 = 1.96 gm3. Volume of concentrated acid needed =

Make 1 Litre of 1M aqueous solution of H2SO4• Read the label on the bottle of H2SO4 reagent. (18M)• 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

Typical concentrations of concentrated acid & Bases (As on Lable)Name Wt% Density (Sp.

Gr)Gm/ml

MWGm/Mole

Molarity

Acetic acid 99.7 1.05 60.05 17.4Ammonium Hydroxide

35

(Aqueous ammonia) 28 0.89 17 14.6Hydrochloric acid 37 1.18 36.46 12Nitric acid 70 1.40 63 15.6Phosphoric acid 85 1.69 14.7Sulfuric acid 96 1.84 98.07 18Density = weight of 1 ml liquid

Shortcuts

• For Molarity, n=1• For Normality, n=calculated

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

Molar solutions from liquid

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

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)

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)

Thanking You