頁二第張五第日五十月三年丑癸夏
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Chemistry (25)
Solution for last week
Part I
(廿五)
Aqueous sodium chloride can
be electrolysed by using the apparatus as shown. At the cathode, sodium ions are dis- charged
•Na +
Na
The sodium metal formed sodium amalgam with mercury. At the anode chloride ions are preferentially discharged.
C1
01
*
3.,
51.
Part II
10. Q
Ratio of weights of H:0:C:Ag
0.0143 0.114: 0.449: 1-542 17.97 31:40:107.6 The ratio of equivalent weights obtained from the atmic weights.
∙107.9
= 1.008 8.0.3 31-75215 of
Hence, within the limits of experimental working; the experimental results agres with Faraday's Second Ian. (ii) Quantity of electricity
passed = 0.65 x 2100 Coulombs =1365 Coulombs.
1365 Coulombs liberate 1.542.
of Silver
1:542
Coulomb. liberate 1365
of silver 0.00113 gm. of silver
(a) An electrolyte is a sub- stance which, in solution or in the molten state, will conduct an electric current and at the same time is decomposed by it.
Electrolysis is the decomp- osition of a compound (molten or in solution) by the-passage through it of an electric current. My d
The anode is the positive electrode by which electrons leave an electrolyte..
The cathode is the negative electrode by which electrons enter an electrolyte.
pole
of battery
Hydrofzin.
water found
little and do 504
batting.
(b) During the electrolysis of water, no SO ions are discharg -ed and sulphuric acid in the solution is not altered anolys- ing the solution before and after electrolysis. The norma- lity of the dilute sulphuric acid is carefully determine before experiment. The volume of the solution used for the electrolysis is accurately. measured: The amount of sulphuric acid in the solution is therefore known. During the electrolysis water is slow Ly decomposed by current into: oxygen and hydrogen, and the amount of water present gets .less and less so that the acid- ity of the solution increases. After electrolysis the volume- of the solution left and its normality are acourately deter mined. The amount of sulphuric acid is therefore calculated. It is found the same weight of sulphuric acid is left at the end as was added at the begin.
(a) Nothing is seen during the whole process. But the weight of the electrodes is changed. The decreased weight of posit-. ive electrode(the anode) is. equal to the increased weight. of the negative electrode. The fact can be illustrated by: the following experiment. The electrodes are.
and dried.Plym
aned
ii) The electrodes are weighed
separately and each of their weights are recorded iii) The electrodes are dipped
into the copper sulphate solution and a current is: passed through about 20 minutes.r
iv) The electrodes are cleaned
and dried and weighed again. It will be found that the lost weight of anode is equal to the increased weight of the cathode
Kleitaskujen of Scrbum chloresta Salution
Brime
Quantitative work
The Molarity and standard solution.
If one mole 1.e. the formi la weight in grammes, of a substance is dissolved in. water and the solution made up to one litre, such a solut ion called a molar solution.
A standara' solution is one
containing a known weight l
solute in a known volume.
We shall use the following relationships in our calculat-.
ions:
Molarity
grammes per litrẻ
molar weight
grammes per
Molar weight = Litre
molarity
Grammes per litre -Molar weight x molarity. These mo
molar
weights wi11.
frequently be used.
Hydrochloric acid
Nitric acid
sulphuric acid
acetic acid
sodium hydroxide".
40
potassium hydroxide
56:
sodium carbonate
sodium bicarbonate.
Titration
In determining the molarity of an acid it has be titrated against a muitable base. example
For
To prepare an exactly 0.1M stand- ard solution of sodium carbonate. Step 1: Alkaline solution:
Weigh out accurately 2.65 gm. († nol. wi.) of anhydrous sodium carbonate on a watch glass, and dissolved it to the warm water in e beaker. Transfer the cooled solution from the beaker into the 250 cm3
volumetric flask. Finally. make up to the mark of the volum metric flask with distilled water. Step 2: Acidic solution:
Conc. sulphuric acid (S.G. 1.84)
The mol. wt. or H 50, is 98gm
3
250 om 0.1M
requires
the
.4.gm is equivale
ution
Measure out the acid(1.4 cm3). with a small graduated cylind -er and pour it into a 250.-
13 flask. Add distilled water and make up the mark of the flask.
сд
3: Standardisation... The acid, in a burette, is added to exactly 250 cm of. standard sodium carbonate in a conical flask. Add two to three drops of methyl orange to the conical flask.
Add the acid, 1 cm at a time, to the flask. Add acid until the indicator changes colour. Read the burette. Calculate the volume of acid used. Repeat the titration with the carbonate in the second and third flasks."
If: X moles of acid react with
Y moles of alkali;
Molarity X Volume of acid Molarity X Volume of alkali
Dilution
To prepare exactly 0.1M hydro-
acid (500 cm3) from 0.1.08M
chloric acid
Molarity X volume.
of dilute. acid
Molarity I Volume of ori ginal acid.
0.1 X Volume of dilute acid
0.108 x 500 cm2
.e. volume of dilute acid
0.108 x 500 cm3. -540 cm3
0.1
Therefore add distilled water. to 500 cm3 of the 0.106M acid until the volume of solution is
540 cm. cm3.
Indicators:
Indicators are substances which vary in colour according to the Hion concentration of the solut- ion or liquid to which they are added. Different indicators change colour over different ranges pH. Some of the commoner indicators. are listed below:
im
Co
Indicator
Bromothymol blue yellow
phenolphthalein
Litmus
Cresol red
less
colour...
日七十月四年三七九一膰公年二十六國民華中育教僑華
It can also be calculated in
the following way:
In an acid-alkali titration, is important in choosing an in- dicator which will give the correct end point to pick one which changes colour over the range in which there is a marked pH change during the titration. The ideal indicator would change colour over a range whose mid- point was the mid-point of the marked pH change occurring during
the titration,,
Indicator
such as
strong alkali Strong acid and
Titration
Marked pH
change
strong alkali 7-5 - 10.5 Weak acid and
Strong acid and weak alkali:
Weak acid and No marked
weak alkali
change
almost any indicator Litmus (58) or
phenolphthalein (8.3-10)|
methyl red (4.4 - 6.3)
detected accurately by End point can not be
any indicator.
10
pH values
3.5-6.5
· The equilibrium between nole- cular water and its ions may be represented by.
H
[H] x [OH]
OH
Here again, [H0] is consider- ed constant and is not written as one of the variable terms in the equilibrium constant equation. Because of the great importance of H0 and its ions, the numical value of this equilibrium constant should be remembered. At. 2500 the value of the water constant,
K = [1') x (OH) is 1.0 x 10-4 In pure water [H] = [OH]
1.0 x107
14:
mole/liter,
1.0 x 10-7
In acid solution is one in which the (H') is greater then 10-7
mole/Liter,
basic solution is one in which
the (H) is less than 10 mole/
liter..
The acidity or alkalinity of
a solution is expressed often by the common logarithm of the reci roes of the hydrogen ion concent- ration in mole per liter. This value is called the pH of the solution. Thus by definition.
pl = 106 → = −106 [H*]
or [1'] = 10 PH For example, if [x]
then pH = 10g
log 102
10g.
Similarly, pOH = (OH) and pH + POH
--14
For example, if the pH of a solution is 2.6, its pOH 14-2.6
11.4
Calculation examples
25.om of 0.1M NaOH neutralize 21 cm3 HCl. Find the concent- ration of the later and calcu late how much water must be added to a litre of it is order to make it exactly decimolar.
→ NaCl + H20 NaOH + HCL
25 x 0.1M(alkali) = 21 x M(acid.
M0.119M
Grammes per litre molarity
x molar weight 0.119 x 36.5 4.34gm
If 40 cm3 of
are added tc each 21 cm3 of the acid the solution will become exactly decimolar as then the neutrali- zing volumes will be equal. Therefore, the amount of water to make one litre of the solut- ión decimòlar will be:
4 x 1000
=190 cm3
1000.cm
of 0.119M HO1
Vem of 0.1M HC1
from which
V = 1190 cm3
the water to be added
190 cm3
If 0.48 gm. of a divalent metal dissolves in a dilute acid with
the liberation of 192 cm of hydrogen. What is the atomic weight of the metal.
•X012 + H2
X + 2HC1
22.4 litres of hydrogen are liberated by 1 g-atom of the metal X
192 cm of hydrogen are.
liberated by
atom of X
·192 22400
g-aton
As this fraction weighs 0.48gm.
gu-atom must weigh 0.48 22400
192
gm. 56 gm
What volume of carbon dioxide will be liberated at 20°C and 745 mm pressure by excess molar H,50 from 0.86 gm. of Na CO? What volume of acid
4.
will be used up?
Na2CO3
mole of Na2 CO (106 gm) oduces 1 mole of CO2 (22.41)
0.86
mole Na co
will:
0.86 x 22.4) CO2
produce T06
at
182 cm
At 2000 and 745 mm pressure this volume will be
293
760 l - 199.2 cm 745
mole of CO2 requires i male
H250 for its prod ation 10.86
0.86
106 mole requires mole
of acid.
This amount be present in
100 x
Exercise
Part I
0.86
106 m3
volume of acid
-8.1 cm
Given that [H] (OH)
= 10-14
the pH of a 0.1M solution, of potassium hydroxide is.
A. 8
B.. 10:
E. 13:5
Which of the following substan- ce in 0.1M aqueous solution would give the highest pH?
sodium chloride
B. Ammonium acetate
C. Hydrochloric acid D. Ammonium nitrate"
B. potassium acetate
When titrating sodium hydroxide. against acetic acid which of the following indicators are suitable?
A. phenolphthalein.
B. methyl orange
C. Screened methy, orange Dimethyl red
E. Litmus
1. dm of molar oxalic acid
reacts with 2 dm3 of molar sodium hydroxide. From this it follows that
2. am of 0.5M oxalic acid
3. react with 1 dm of 2M sodium hydroxide.
0.5 dm3 of 2M oxalic acid.
react with 4 dm of 0.5M sodium hydroxide.
C. 0.5 dm3 of 2M I oxalic acid
react with 1 dm of M sodium hydroxide.
D. 0.5 dm of 0.5M oxalic acid. react with 0.25 dn of M sodium hydroxide.
2
dm3 of 0.5M oxalic acid
react with 0.5 dm3 of M. godium hydroxide.
5. A solution contains 20 gms of
potassium hydroxide per litre. 20 cm of this solution were completely neutralized by 40cm of an acid of concentration 15 gm/litre. What is the equivalent of the acid?. A. 3:1
B. 15 0.30
D. 45 E.. 60
Part II
15
Calculate the (a) molarity and (b) molarity of a sulphuric acid solution of specific gravity 1.198, containing 27%
by weight.
cm of 0.1M HO1 neutralize 15 cm3 of a solution of sodium carbonate; find the molarity of the latter.
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