Laboratory Solution Preparation

Prepared:

Farhang Hamid

ئاواددةكسدنى طرياوة لة تاقيطةدا(Laboratory Solution Preparation)

ئاواددةكسدنى طرياوة لة تاقيطةدا طسنطييةكى تايبةتى خؤيى يةية بؤ تاقيكسدنةوةكاى وةك تاقيكسدنةوة لةسةز خؤزاك ياى ياى ئاواددة بكسيَت,ئاواددةكسدنى طرياوةش كؤوةلَيَك تاقيكسدنةوةى وايكسؤبايؤلؤجى ياى كيىيايى وة ئةم طرياوانة ياى ثيَويستة بكسِديَت

ةزِووى سةالوةتى و باشى طرياوةكة,طرياوةكةش خةستى يةية ئةم خةستيةش لةسةز بهةواى تواوة و تويَهةز دةبيَت .وةزج و بهةواى يةية,ل

-يةنديَك لة بهةواسةزةكييةكانى ئاواددةكسدنى طرياوة:

(Molarity):-وؤالزيتى

يةكيَكة لة بهةواسةزةكييةكانى يةكةى طرياوة كة ناويَكى باوة بؤ خةستى طرياوةكاى بةكازديَت ,وؤالزيتى بسيتيية لة ذوازةى وؤالزيتى .ة ليرتى طرياو 1ةكانى تواوة لةسةز وؤلَ

-بؤواددة زِةقةكاى ثيَويستىاى بةم ياساية دةبيَت:

( ) ( )

Ex/ Prepare 800 ml of 2 M sodium chloride NaCl M.wt(NaCl)=58.45 g/mol

Weight=M.wt(s) × M × V

Weight= 58.45 × 2 × 0.8=93.52 g NaCl

ون ئاوى دلَؤثيَهساو دةيتويَهيهةوة دواتس تةواوكةزةكةى بؤ شياددةكةيو 400طسام لة كمؤزيدى صؤديؤم سةزةتا لة 23.52

-ئةم ياساية بةكازدةييَهني: بؤ واددة شمةكاى

M1×V1=M2×V2

Ex/ Prepare 100 ml of 1.0 M hydrochloric acid from Concentrated (12.1 M) hydrochloric acid.

M1×V1=M2×V2

12.1 × V = 1.0 × 100

ون ئاوى دلَؤثيَهساو , تيَكةهَ دةكسيَت و ثاشاى تةواوكةزةكةى بؤ شياددةكةيو. 50شياددةكةيو بؤ HClون لة 6.26

Percent Solution طرياوة زِيَرةييةكاى

Mass percent Solution:-

( )

( )

Ex/ 20g of sodium chloride in 100 gram of solution.

Volume percent Solution:-

( )

( )

Ex/ 10ml of ethyl alcohol plus 90 ml of H2O (making approx..100 ml of solution) is a 10%

by Volume solution.

Mass – Volume percent Solution:-

( )

( )

Ex/ 1 gram phenolphthalein 100 ml of 95% ethyl alcohol is a 1 w/v% solution.

Calculating Molarity from percent Solution

يةذوازكسدنى خةستى لة طرياوة زِيَرةييةكاى

سةزةتا دؤشيهةوةى كيَشى طرياوةكة لة زِيَطةى ئةم ياسايةوة -1

%بؤ كيًش دواتس طؤزِيهى % بؤ ئاسايى واتة كيَش -2 0.362دةبيَت بة 36.2من/ %

ى طرياوةكة )دزاوة( M.wtديازيكسدنى -3

دؤشيهةوةى وؤلَى طرياوةكة -4

.لةكؤتايدا خةستى دةدؤشيهةوة -5

Ex/ Determine molarity of 37.2% hydrochloric acid (density 1.19 g/ml)

Mass%=37.2=0.372 gram

M.wt=36.4 g/mol

Concentration units

1-Normality of solution:-

Is defined as the number of grams-equivalent weight of solute per one liter of solution .

2- Molarities of solution :-

A molar solution contains one gram molecular weights of solute dissolved in sufficient

amount of solvent to give(1000ml) 1 liter of solution .

V (ml)

= XM =gram molecular wt. of solute

volume of solvent (L)

wt

MWT 1000

Formality :- is the number of grams formula weight of solute per liter of solution. used

for partial dissolved solutes (such as acetic acid , oxalic acid etc…)

Preparation of 1% w/v Na2CO3 solution

Weigh 1 gm of sodium carbonate using a watch glass, transfer it into a beaker, then add

50 ml. of distilled water(mix using a stirrer )transfer the solution into a volumetric flask of

100 ml. add distilled water to the mark.

Mix the solution by inverting the flask 2-3 times to mix the component .

Q / prepare 0.3 % solution from 1% solution in 250 ml volumetric flask.

Preparation of 0.1N Na2CO3 solution

Weigh 1.325 gm of Na2CO3 (using a watch glass ) transfer it into a beaker. Add 100 ml

of distilled water (using a stirrer to mix), after dissolving , transfer the solution into a

volumetric flask of 250 ml then add water to the mark .

Q / From this solution prepare 0.05 N , and 0.02 N solution in 250 ml distill water ?

Wt. = eq wt. x N X V (ml)

1000

To prepare dilute solution from stock solution , this law must be used :-

Dilution law is :- (N x V)1 = (N x V)2

Where (1) refer to the concentrated solution and (2) refer to the diluted solution.

To find the normality of concentrated solutions (for liquid only) this law must be used:-

N = density or spicific gravity x wt % x 1000

eq.wt of solution

Or :

density or spicific gravity x wt % x 1000

M =

MWt. of solution

Q / prepare 2N solution in 250 ml volumetric flask from 5N of HCl stock solution .

Q / prepare 500 ml , 5N H2SO4 solution from original concentrated solution. .

Preparation of 0.1 M Na2CO3 solution in 250 ml D.W

Weight 2.65 gm of Na2CO3 , transfer it into a beaker. Add 100ml of D.W ( mix using a

glass stirrer ).Transfer the solution into a 250 ml volumetric flask of, add water to the

mark.

ppm = wt of solute x 10 6

wt of solution

ppm = mg (solute)

10 6 mg water

mg (solute)

L (solution)=

mole of solute = Wt (gram)

Molecular weight (g / mole)

eq.wt of solute =

Molecular weight (g / mole)

no of H + In acids

no of OH - in base

no of cation charge in salts

So : 1M Na2CO3 = 2N Na2CO3

part per million (PPm )

The concentration of very diluted solutions ,It’s the number of mg of salt dissolved in

one liter solution.

ppm = wt of solute x 10 6

wt of solution

ppm = mg (solute)

10 6 mg water

mg (solute)

L (solution)=

Buffer solution

Any solution that contains both a weak acid and it’s weak base, has the ability to absorb

small amounts of either a strong acid or strong base are added they are neutralized by

the weak base , while small quantities of a strong base are neutralized by the weak acid,

such solution are said to be buffer because they resist significant change in the PH .The

PH value is determined by a PH meter.

Using of indicators in a PH estimation:-

Principle:-

A set of buffer solution of known PH is prepared form citric acid and disodium

monohydrogen phosphate.

The color of the indicator in an unknown solution is compared with the color of indicator

in these buffer solution , assuming that the concentration of the indicator in all solution is

the same.

The buffer solution in which the color of the indicator is the same as color of the indicator

in the unknown solution has the same PH as this solution.

Procedure :-

Fill one burette with citric acid solution 0.1 molar and a second burette with 0.2 molar

disodium monohydrogen phosphate ( called solution A and B respectively) .

Use the following table in preparing a set of buffer solutions:-

No PH ml of solution A ml of solution B

1 4.2 5.86 4.14

2 4.4 5.59 4.41

3 4.6 5.33 4.68

4 4.8 5.07 4.93

5 5 4.85 5.15

6 5.2 4.64 5.36

7 5.4 4.43 5.58

8 5.6 4.2 5.8

9 5.8 3.96 6.03

10 6.0 3.69 6.32

The way to work is follows:-

1- Label 10 test tube from 1 to 10 , into number 1 run 5.86 ml of solution A and 4.14

ml of solution B , mix well, fill the other test tube in a similar way , using the

volumes of A and B mentioned in the table. In this way we obtain a set of buffer

solution (10 ml of each ) having a PH from 4.2 to 6.0

2- Use methyl red as an indicator ,add to each 10 ml buffer solution above, 5 drops of

the methyl red indicator solution .(PH rang 4.4 to 6.0 ),mix the contents of the

tubes well.

3- Bring into a test tube 10 ml of a 0.05 N ammonium chloride solution ,add 5 drops

of the indicator solution and mix .compare the color produced with the color in the

test of buffer solution.

4- Using a pipette bring into a flask 10 ml 0.1 N acetic acid solution and 10 ml 0.1 N

sodium acetate solution .mix then ,bring 10 ml of this buffer solution into test tubes

,add 5 drops indicator solution ,and make an estimation of PH of this buffer

solution as before.

Preparation of a buffer solution with PH = 4.6

Mix 5 ml of 0.2 M acetic acid with 5 ml of 0.2 M sodium acetate.(mix using a stirrer).

Preparation of a buffer solution with PH = 7

Mix 17 ml of 0.02 M Na2HPO4.2H2O with 3.9 ml of citric acid .

Preparation of a buffer solution with PH = 9.4

Mix 5 ml of NaOH , 0.1N with 16.6 ml of 0.15 M Na2HPO4.2H2O