practical ( 2). we have two ways to prepare solutions according to the nature of substances

SOLUTION PREPARATIONSOLID SOLUTIONLIQUID SOLUTI

Practical ( 2)

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

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

Important laws

M = n / V ( L )n = m / M.wtM = 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

Example :To prepare ( 0.02 M ) from sodium hydroxide solution NaOH in 250 ml ?

M.wt for NaOH = 23 + 16 + 1 = 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.

Experiment ( 1 )

Objective :Prepare sodium carbonate as a standard solution to measure another substance of unknown concentration

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.

Materials:

1. Sodium carbonate solid 2. Volumetric Flask 100 ml 3. Filter paper 4. Balance5. Small beaker

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.

Liquid solution Dilution of Solutions: To prepare dilute solution from concentrated

one, we need to add more solvent to the concentrated solution.

Dilution of Solutions

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

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 = 1 + 35.5 = 36.5 g/mol

Mconc = P × D × 10 = 38 × 1.18 × 10

M.wt 36.5

= 12.28 mol/L

From the Dilution law Mdil × Vdil =Mconc × Vconc

0.1 × 50 ml = 12.285 × Vconc

Vconc = 0.407 ml

Very important note: To solve acid in water we must add

acid to water not reverse.

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).

Experiment ( 2 )

Objective: Prepare a standard solution of

hydrochloric acid HCl ( 0.01 M ).

Material :

1. Hydrochloric acid solution. 2. Volumetric flask 100 ml.3. Small beaker. 4. Graduated Pipette.5. rubber suction bulb.

procedure:

1. Rinse material with distilled water .2. 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

3. Put an amount of distalled water in the volumetric flask then add a certain volume of HCl to the water in the flask .

4. Fill up with distalled water to the mark.5. Keep the solution till do the titration.

b0

Thank you