1. Solution preparation Solid solution Liquid soluti
Practical # 1
Solution preparation Solid solution Liquid soluti

2. We have two ways to prepare solutions according to the nature of substances.

3. Solid state :
To prepare solution from solid, we must know the concentration which will be prepared in a certain volume.

4. Important laws
M = n / V ( L ) n = m / M.wt M = m / M.wt × V ( L ) m = M × M.wt × V ( L ) where M : molarity n : mole number V : volume in litter M.wt: molecular weight m : weight in grams

5. Example : To prepare ( 0.02 M ) from sodium hydroxide solution NaOH in 250 ml ?
M.wt for NaOH = = 40 g/mol
m = M × M.wt × V ( L )
m = 0.02 × 40 × 250 /1000 = 0.2 g
So...
0.2 g needed to be solved in 250 ml of distilled water to get solution 0.02 M of sodium hydroxide.

6. Experiment ( 1 )
Objective :
Prepare sodium carbonate as a standard solution to measure another substance of unknown concentration

7. Give the reason :Sodium Carbonate Na2CO3 is standard solution
Sodium Carbonate Na2CO3 is a standard solution because it is easy to get it pure, its character don’t change by weighting in air . It has big molecular weight. It can be easily solved in distilled water. It reacts fast with other substances.

8. Materials : Sodium carbonate solid Volumetric Flask 100 ml
Filter paper
Balance
Small beaker

9. Procedure / Method: Rinse materials with distilled water .
Weight Sodium Carbonate Na2CO3 by using filter paper to prepare 0.04 M
m = M × M.wt × V ( L )
Dissolve the amount of the solid with little amount of distilled water in a small beaker to make solution.
Add solution in a volumetric flask and continue with distilled water till the mark.
Keep the solution till the titration.

11. Liquid solution Dilution of Solutions:
To prepare dilute solution from concentrated one, we need to add more solvent to the concentrated solution.

12. Dilution of Solutions

13. Mdil × Vdil =Mconc × Vconc
Dilution of Solutions
Since moles are constant before and after dilution, we can use the following formula for calculations.
Mdil × Vdil =Mconc × Vconc
To prepare a diluted solution from a liquid concentrated one, you must know the molarity for the concentrated solution.
density ( D )
purity ( P )
M = P × D × 10
M.wt

14. Example:
How can we prepare 50 ml of ( 0.1 M ) of hydrochloric acid HCl from a concentrated HCl with P = 38% & D = 1.18 Kg/L
Solution:
M.wt = = 36.5 g/mol
Mconc = P × D × = × 1.18 × 10
M.wt
= mol/L

15. From the Dilution law Very important note: Mdil × Vdil =Mconc × Vconc
0.1 × 50 ml = × Vconc
Vconc = ml
Very important note:
To solve acid in water we must add acid to water not reverse.

16. Standard solution :
Any solution which has a precisely known concentration.
Or in another word:
It is a solution that can be prepared by dissolving a certain weight in a certain volume.
Standard solutions are used to determine the concentrations of other substances, such as solutions in titrations.
The concentrations of standard solutions are normally expressed in units of moles per liter (mol/L).

17. Experiment ( 2 ) Objective:
Prepare a standard solution of hydrochloric acid HCl ( M ).

18. Material : Hydrochloric acid solution. Volumetric flask 100 ml.
Small beaker.
Graduated Pipette.
rubber suction bulb.

19. procedure: Rinse material with distilled water .
poor suitable volume of concentrated hydrochloric acid HCl in a small beaker, then pipit a suitable volume of it to prepare (0.01 M) of HCl
Put an amount of distal water in the volumetric flask then add a certain volume of HCl to the water in the flask .
Fill up with distal water to the mark.
Keep the solution till do the titration.

20. b0
Thank you