Chemistry Study Material: Solutions, Bronsted-Lowry Theory, Volumetric Analysis, Back Titration

This study material covers topics like solutions, Bronsted-Lowry theory, volumetric analysis, and back titration. It includes equations, calculations, and procedures for conducting experiments.

Chemistry
1. Solutions
a) Dissociation Equation for Aluminum sulfate in water.
A l2 ( S O4 )3 (s )→ 2 A l+3 ( aq ) +3 S O4
−2 (aq)
b) Mass of Aluminum sulfate needed to prepare 200 cm3 of 0.450 mol dm-3
M r ( A l2 ( S O4 )3 )=2 ( 27.0 ) +3 ( 32.0 ) +12 ( 16.0 ) =342.00
Mass ¿ make 0.450 mol∈1dm3=342.00 × 0.450=153.900 g
¿ 200 c m3 , the massis 200
1000 ×153.900 g=30.780 g
c) Concentration of :
i) A l+3, the equation in part a), 1 mole of A l2 ( S O4 ) 3gives 2 moles of A l+3 , thus the
concentration of A l+3is:
2 ×0.450=0.900 mol dm−3
ii) S O4
−2, the equation in part a), 1 mole of A l2 ( S O4 ) 3gives 3 moles of S O4
−2 , thus the
concentration ofS O4
−2is:
3 ×0.450=1.350 mol dm−3
d) If 100 cm3 of 1 mol dm-3 sodium Sulfate in the solution in b) above, then, the
concentration of:
i) A l+3, Number of moles of A l+3in 200 cm3 of solution is:
0.900 mol
dm3 × 200
1000 d m3 =0.18 0 mol
In 300 cm3 (total resulting volume), then the number of moles in 300 cm3 is:
0.180 mol × 300 c m3
200 c m3 =0.27 0 mol

Then the concentration is:
0.270 mol × 1000 c m3
300 c m3 =0.900 mol /d m3
ii) Na+¿¿, As from part i)the concentration of Na+¿¿ is
2 ×1.0 mol d m−3=2.000 mol d m−3
iii) S O4
−2, Number of moles ofS O4
−2, in 200 cm3 of solution is:
1.350 mol
dm3 × 200
1000 d m3=0.270 mol
Number of moles of S O4
−2, in 100 cm3 of 1 molar solution is:
0.100 mol
dm3 ×1 d m3=0.100 mol
Thus, total moles of S O4
−2is 0.37 mol in 300 cm3
The molarity is therefore;
0.370 mol × 1000
300 d m−3 =1.233mol / d m3
2. Solution.
a) When aqueous solution of AgN O3 ( aq) is mixed with KCl.
AgN O3 ( aq )+KCl ( aq ) → KNO3 ( aq )+ AgCl( s)
i) A white precipitate is observed when AgN O3 ( aq) is mixed with KCl
ii) The full ionic equation is:
Ag+¿ ( aq )+ N O3
−¿ (aq )+ K+¿ ( aq ) +Cl−¿ (aq ) → K 3
¿ ¿ ¿ ¿ ¿¿
iii) The net ionic equation is:
Ag+¿ ( aq ) +Cl−¿ ( aq ) → AgCl(s )¿ ¿
b) Sodium sulfide react with silver chlorate as given below:

2 AgClO3 (aq)+ Na2 S (aq) → Ag2 S (s )+2 NaClO3 (aq)
Full ionic equation is:
2 Ag3
¿ ¿
Net ionic equation
2 Ag+¿ ( aq ) +S−¿ ( aq ) → Ag 2S ( s ) ¿¿
Both Sodium sulfide and silver chlorate are soluble salts, when they react an insoluble
salt silver sulfide is formed
3. Bronsted-Lowry theory
a) An acid is defined by Bronsted-Lowry theory as any ion or molecule which is capable of
donating a hydrogen cation, H+¿¿, (proton) (Kolthoff, 1944)
b) Conjugate acid is a species formed by the reception of a proton by a base, while
conjugate base is the species that has left over after the acid has donated a proton during a
chemical reaction. That is:
Base+ Acid ⇌ conjugate acid+conjugate base
For instance:
H2 O ( l ) + HCl O2 ( aq ) ⟶ H3 O+ ¿ (aq )+ClO 2
−¿(aq) ¿¿
Where H2 Ois base, HCl O2 is acid, H3 O+ ¿¿is conjugate acid and Cl O2
−¿ ¿is conjugate
base.
c) In aqueous solution :
i) Equation on how H2 O behave like an acid
N H3 ( aq ) + H2 O ( l ) → N H4
+¿ ( aq )+O H −¿ (aq ) ¿ ¿
ii) Equation on how NO3
−¿¿ behave like a base
N O3
−¿ ( aq ) + N H 4
−¿ ( aq ) →HN O 3 ( aq ) + N H 3 ( aq ) ¿¿

iii) Equation on how NH 4
+¿¿ behave like an acid
N O3
−¿ ( aq ) + N H 4
−¿ ( aq ) →HN O 3 ( aq )+ N H 3 ( aq ) ¿¿
iv) Equation on how H2 O behave like a base
HF ( aq )+ H 2 O ( l ) → F−¿ (aq )+ H3 O+ ¿ ( aq ) ¿¿
d) In the given reactions
i) C H3 COOH ( aq ) + HClO4 ( aq ) →C H 3 COO H2
+¿ ( aq ) +Cl O4
−¿ ( aq ) ¿¿ HCl O4 ( aq ) ac t like acid
C H3 COO H 2
+ ¿is conjugate acid ¿Cl O4
−¿ (aq ) is conjugate base ¿
ii) H2 S O4 ( aq ) + HN O3 ( aq ) → HSO4
−¿ (aq )+ H2 N O3
+ ¿ ( aq ) ¿¿ HN O3 act as an acid
HS O4
−¿is conjugate base ¿ H2 N O3
+¿ isconjugate acid ¿
4. Volumetric analysis
a) Volumetric analysis is quantitative analytical method used in measuring volume of a
solution of known concentration that is used to determine the concentration of the analyte
(Taleuzzaman & Gilani, 2017).
b) How to determine the exact concentration of nitric acid.
 Apparatus requires to perform the experiment is:
Phenolphthalein indicator, 50 cm3 nitric acid, 100 cm3 of 0.1 Molar potassium hydroxide,
Burette, Clamp and Stand, Volumetric flask, Pipet and a beaker
 Procedure
i) Wash and rinse all the apparatus using distilled water
ii) Using pipet, measure exactly 25cm3 of nitric acid and pour it in volumetric flask.
iii) Add 2-3 drops of phenolphthalein in the flask, which is colorless when in acid.

iv) Clump the burette with stand and clump, and fill the burette with 50 mL of KOH and
let two drops flow through to clear any air bubbles into the beaker
v) Remove the beaker and put the volumetric flask containing the acid
vi) Carefully, add a drop of KOH in the acid, continue adding the drops till the color of
the indicator start acid turns a very light pink. At this point the acid has completely
reacted with the base. (This is the equilibrium point)
vii) Record the volume of base used to react the acid.
viii) Perform calculations (done in part C below)
c) Assuming the volume of 0.1Molar potassium hydroxide used is 34.6 cm3.
i) The balanced chemical equation for the reaction
KOH ( aq )+HN O3 ( aq ) → KN O3 ( aq ) + H2 O(l)
ii) Number of moles of KOH reacted
nKOH= 0.100 ×34.6
1000 =0.00346 mol
iii) From the reaction equation, the number of moles in 25 cm3 of HNO3 is 0.00346
moles, since the ratio is 1:1.
iv) The concentration of nitric acid is calculated as:

CHN O 3
= 0.00346× 1000
25.0 =0.1384 mol d m−3
5. Back titration
a) A back titration (also known as indirect titration) is a titration method where unknown
concentration of an analyte can be found by reacting it with excess known amount of the
reagent. Further, the excess reagent remaining is titrated with a known second reagent.
This helps to know how much excess reagent used in the first titration, which allows te
determination of the concentration of the analyte (Helmenstine, 2018).
b) Calculation on basis of the information given the following
100 cm3 of 1.00 Molar NaOH(aq)
i) The initial number of moles of NaOH(aq)
If 1.00 mol is in 1000 cm3, 100cm3 contains:
nNaOH ( aq ) =1.00 mol /d m3 ×100 c m3
1000 c m3 /d m3 =0.10 0 mol
ii) The excess amount of NaOH(aq) that reacted with H2SO4(aq) is calculated as
follows:
2 NaOH ( aq ) + H2 S O4 ( aq ) → N a2 S O4 ( aq ) +2 H2 O ( l )
nH 2 S O4 ( aq ) = 0.0250 ×50.0
1000 =0.00125 mol nNaOH ( aq ) ( excess ) =2× 0.00125=0.0025 mol
iii) The amount of NaOH that reacted with NH4Cl(aq) equal to total amount of
NaOH(aq) minus the excess NaOH that reacted with H2SO4(aq)
nNaOH ( reacted with N H 4 Cl ) =nNaOH ( aq ) −nNaOH ( aq ) ( excess ) ¿ 0.100−0.0025=0.0975 mol
iv) Amount of NH4Cl(aq) in the sample is:

From the equation,N H4 Cl ( aq )+ NaOH ( aq ) → N H3 ( g )+ NaCl ( aq ), the ratio of
NH4Cl(aq): NaOH(aq0 is 1:1, thus the number of moles of NH4Cl(aq) in the
sample is 0.0975 mol
v) Mass of ammonium chloride in the sample
M r ( N H4 Cl ) =14.0+4 ( 1.0 ) +35.5=53.5
mass=Mr ( N H 4 Cl ) ×nN H 4 Cl =53.5× 0.0975¿ 5.2168 g

References
Helmenstine, A. M., 2018. Back Titration Definition. [Online]
Available at: https://www.thoughtco.com/back-titration-definition-608731
[Accessed 21 January 2019].
Kolthoff, I. M., 1944. The Lewis and the Brönsted–Lowry Definitions of Acids and Bases. The
Journal of Physical Chemistry, 48(1), pp. 51-57.
Taleuzzaman, M. & Gilani, S. J., 2017. First Step Analysis in Quality Control -Volumetric
Analysis. Global journal of Pharmacy & pharmaceutical Science, 1(3), pp. 001-003.