Preparation of standard, normal and molar solutions

Preparation of standard, Normal and molar solutions
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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 10 23 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 10 23 mol¯ 1 Problem #2: 0.200 mole of H 2 O contains how many molecules ? 0.200 mol x 6.022 x 10 23 mol¯ 1

4. 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

5. Calculate the number of moles in 1.058 gram of H 2 O. (MW of H 2 O=18 gm /mole) Calculate the number of molecules in 1.058 gram of H 2 O  

6. Determination of molecular weight ( H 2 SO 4 ) 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

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

8. 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 H 2 SO 4  1 mole of H 2 SO 4 in 1 litre of solution But it gives 2 moles of acid So, 1M of H 2 SO 4 will be equivalent to 2 moles of acid

9. 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 H 2 SO 4 = 4N 2M H 3 P O 4 = 6N 2M HCl = 2N 2M NaOH = 2N

10. Ideal procedure of calculating Normality 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 H 2 SO 4 ’s equivivalent mass = 98/2 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

11. If compound is liquid then follow below formula   Prepare 200 ml of 0.2 N H2SO4 solution MW = 98, number of H+=2 so eq.wt =49 Gms of Compound needed = 0.2 x 49 x 0.2 = 1.96 gm Volume of concentrated acid needed =  

12. 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  

13. 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  

14. Shortcuts For Molarity, n=1 For Normality, n=calculated  

15. 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

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

17. 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)

18. 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)

19. Thanking You