Topic 4 - Acids and Bases (Titration).pdf

N/A

1. Steps in titration
Finally,takeareasonablevolumeof
solutionBrequiredintitration

2. The conical flask used in titration has to be cleaned by washing with
water. (It should not be washed with the solution it is to hold.) The
washed conical flask need not be dried before use. Explain.
•The conical flask is to hold a specific volume of a solution (usually
25.0 cm
3
) and hence a specific amount of the solute. The conical flask
has to be washed with water to get rid of impurities, which may affect
titration results. It should not be washed with the solution, since an
additional amount of solution would certainly affect titration results.
• After washing, a little water would remain in the conical flask.
This will not change the amount of solute present. Thus we should
not waste time in drying the conical flask.

3. Cleansing by washing with water/ solution to be contained?
•Get rid of impurities? ( Distilled water should be used.)
•Additional amount of solution would certainly affect titration results?
(i.e. Will affect the total amount of solute present in the container,
BUT IS IT IMPORTANT?)

4. End Point Detection in Acid-Alkali Titrations
Titration type Example Suitable indicator(s)
Strong acid -strong alkaliHCl(aq) vs. NaOH(aq) methyl orange;
phenolphthalein
Strong acid -weak alkali
(Strong acid -carbonate)
HNO3(aq) vs. NH3(aq)
HCl(aq) vs. Na2CO3(aq)
methyl orange
Weak acid -strong alkaliCH3COOH(aq) vs. KOH(aq) phenolphthalein
Weak acid -weak alkaliCH3COOH(aq) vs. NH3(aq)no suitable indicator
1. By use of indicators
An Indicatoris a substance which changes colour within a specific pH range. Commonly used indicators
are methyl orange and phenolphthalein..

5. End Point Detection in Acid-Alkali Titrations
Titration type Example Suitable indicator(s)
Strong acid -strong alkaliHCl(aq) vs. NaOH(aq) methyl orange;
phenolphthalein
Strong acid -weak alkali
(Strong acid -carbonate)
HNO3(aq) vs. NH3(aq)
HCl(aq) vs. Na2CO3(aq)
methyl orange
Weak acid -strong alkaliCH3COOH(aq) vs. KOH(aq) phenolphthalein
Weak acid -weak alkaliCH3COOH(aq) vs. NH3(aq)no suitable indicator
Name a suitable indicator for titration of each of the following complete neutralizations:
a. H2SO4(aq) vs. K2CO3(aq)
b. H2SO4(aq) vs. KOH(aq)
a. methyl orange
b. methyl orange or phenolphthalein.

6. End Point Detection in Acid-Alkali Titrations
2. By use of pH meter
pH meteris an instrument which gives direct reading of pH on the scale 0 to 14.
When 0.1 M HCl is titrated against 25 cm
3
of0.1 M NaOH, the following graph would be obtained:

7. End Point Detection in Acid-Alkali Titrations
3.Byconductivitymeasurements
Changingconcentrationofhydrogenorhydroxideionsinasolutioncausesachangeintheconductivityofthat
solution.Asacid-basetitrationsinvolvechangeofionicconcentrations,thusthereactioncanbefollowedby
meansofconductivitymeasurement.
Againletusconsidertheadditionof0.1MHClfromaburetteintoaconicalflaskcontaining25cm3of0.1M
NaOHandnotetheconductivityofthesolution.

8. End Point Detection in Acid-Alkali Titrations
At A: Conductivity high due to presence of large number of mobile OH-ions from NaOH.
A to E : Conductivity falls as mobile OH-are replaced by less mobile Cl-.
At E: The end-point. Conductivity is at a minimumwhen the neutralization is complete.
Na+(aq) + OH-(aq) + H+(aq) + Cl-(aq) → Na+(aq) + Cl-(aq) + H2O(l)
The conductivity at the end-point is not zero as the solution contains sodium and chloride ions.
E to B : Conductivity increases due to the addition of excess mobile H+ from HCl. The graph
rises more steeply than from A to E as H+ ions are more mobile than OH-ions.

9. Additional Info on Conductivity Measurements
•2017/DSE/1B/Q1c
(c) (i) White precipitate forms
(ii) (1) Ba
2+
react with SO
4
2-
to form insoluble BaSO
4.
The concentration of mobile ion decrease.
(2) Extra unreacted H
2SO
4added increases the
concentration of mobile ion (More H
+
and SO
4
2-
)

10. End Point Detection in Acid-Alkali Titrations
4. By thermometric titration
The reaction H+(aq) + OH-(aq) → H2O(l) is exothermic.
There is the greatest temperature rise at end point.
Consider 25 cm3 of 0.1 M NaOH places in a polystyrene beaker together with a thermometer. 0.1
M HCl is added, a definite quantity at a time, the temperature being noted after each addition. The
following graph is obtained:

11. End Point Detection in Acid-Alkali Titrations
A to E : Temperature rises as more heat is liberated by adding more acid.
At E : The end-point. Temperature is maximum.
E to B : Temperature drops as more acid added will not liberate more heat but will cool the solution.

12. Preparing a soluble salt by titration
To prepare NaCl by titration
Step 1: Place a known volume of NaOH in a conical flask and a few drops of acid-base indicator are added.
Step 2: Titrate the NaOH by hydrochloric acid until the indicator change just changes colour(end point is
reached)
Step 3: Repeat the experiment with the same volume of alkali. Do not use any indicator this time. Add the
volume of acid needed for neutralization (as found from the first experiment).
Step 4: Evaporate the reaction mixture slowly to obtain the salt.

13. Calculations on Volumetric Analysis
1. Mole ratio from equation
Problems on volumetric analysis can be solved by using the mole concept.
We have to write a balanced equation for each reaction that takes place. We should then find mole
ratio of reactants from the equations.
Consider two examples:
a. H
2SO
4(aq) + 2NaOH(aq) → Na
2SO
4(aq) + 2H
2O(l)
The mole ratio of H2SO4 to NaOH is 1:2. In other words, 1 mole of H2SO4 required 2 moles of
NaOH for complete neutralization.
b. CH3COOH(aq) + NaOH(aq) → CH3COONa(aq) + H2O(l)
The mole ratio of CH
3COOH to NaOH is 1:1. In other words, 1 mole of CH
3COOH requires 1 mole
of NaOH for complete neutralization.

14. Calculations on Volumetric Analysis
Example
Findnumber of moles of hydrochloric acid required for complete neutralization of 30 cm
3
of 0.50 M ammonia
solution.
Answer
HCl(aq) + NH
3(aq) →NH
4Cl(aq)
no. of moles of NH
3= (0.50M)(0.03 dm
3
) = 0.015 mol.
According to the equation,1 mole of HCl reacts with 1 mole of NH
3
therefore, the no. of moles of HCl required = 0.015 mol

15. Calculations on Volumetric Analysis
2. Applications of volumetric analysis :
To determine :
(1) concentration of a solution (standardization).
(2) number of water of crystallization of a hydrated salt.
(3) composition of a mixture.
(4) charge of ion.
(5) percentage of ammonia in an ammonium salt (or percentage purity of an ammonium salt).
(6) molar mass of a substance.
(7) atomic mass of a metal.

16. Calculations on Volumetric Analysis
Example 1
2.65 g of sodium carbonate (Na2CO3) were dissolved in water and made up to 250 cm3 solution. 25
cm3 of this required 20.0 cm3 of a hydrochloric acid solution for complete reaction. Find the molarity
of the hydrochloric acid.
Answer
Molar mass of Na2CO3 = 106 g
Number of mole of Na2CO3 in 250 cm3 of solution = 2.65/106 = 0.0250
Number of mole of Na2CO3 in 25.0 cm3 of solution = 0.0250 ×(25.0/250) = 0.0025
2HCl(aq) + Na2CO3(aq) → 2NaCl(aq) + H2O(l) + CO2(g)
From the equation, 2 moles of HCl react with 1 mole of Na2CO3,
∴number of moles of HCl in 20.0 cm3 of solution = 0.0025 ×2 = 0.0050
Number of moles of HCl in 1000 cm3 of solution = 0.0050 ×(1000/20.0) = 0.250
∴Molarity of the HCl solution = 0.250 M

17. Calculations on Volumetric Analysis
Example 2
4.0 g of a sample of hydrated sodium carbonate, of formula Na2CO3.nH2O, are dissolved in water
and the solution made up to 250 cm3. Using methyl orange as indicator, 25 cm3 of this solution
require 29 cm3 of 0.05 M sulphuricacid for neutralization. Calculate n, the number of molecules of
water of crystallization, in the sample of sodium carbonate.
Molar mass of Na2CO3.nH2O = (23 ×2 + 12 + 16 ×3 + 18n)g
Number of moles of Na2CO3.nH2O in 250 cm3 of solution = 4.0/(106 + 18n)
Number of moles of Na2CO3.nH2O in 25 cm3 of solution =
According to the equation
Na2CO3.nH2O(aq) + H2SO4 (aq) ⎯⎯→ Na2SO4(aq) + CO2(g) + (n+1)H2O(l)
Number of moles of Na2CO3.nH2O in 25 cm3 of solution
= number of moles of H2SO4 in 29cm3 of 0.05 M H2SO4
= 0.05 ×29/1000 = 0.00145
∴= 0.00145
∴n = 9.4

18. Calculations on Volumetric Analysis
Example 3
0.949 g of a mixture of sodium hydroxide and anhydrous sodium carbonate was dissolved in 100 cm3 of water. This
solution requires 30.0 cm3 of 0.650 M hydrochloric acid for complete reaction. Find the percentage by mass of the mixture.
Let y g be the mass of NaOH, then the mass of Na2CO3 = (0.949 -y) g
Molar mass of NaOH = 40 g
Molar mass of Na2CO3 = 106 g
NaOH(aq) + HCl(aq) ⎯⎯→ NaCl(aq) + H2O(l)
no. ofmoles of HCl required to neutralize NaOH = y/40
Na2CO3(aq) + 2HCl(aq) ⎯⎯→ 2NaCl(aq) + CO2(g) + H2O(l)
no. of moles ofHCl required to neutralize Na2CO3 = 2×[(0.949-y)/106]
Therefore the total no. moles of HCl = y/40 + 2×[(0.949-y)/106]
From the given data, the no. of moles of HCl = 0.650 x 30.0/1000 = 0.0195
y/40 + 2×[(0.949-y)/106] = 0.0195
y = 0.260
∴% by mass of NaOH = 0.260/0.949 ×100% = 27.4%
% by mass of Na2CO3 = 100% -27.4% = 72.6%

19. DSE/2014/1B/Q7
2 marks

20. DSE/2014/1B/Q7
2 marks
What if the question value 3 marks?
Add a point that ‘Rinse the beaker containing the sodium carbonate solution by several times and drain all the
washings into the volumetric flask.

21. DSE/2014/1B/Q7

22. DSE/2014/1B/Q7
Make sure you have the concept that concentrated hydrochloric acid is volatile
→Loss of solute in form of acid mist
→Exact no. of moles of HCl decreases
→Concentration of HCl(aq) is less than the expected value.

23. Back Titration
Back titration
Back titration is a method used to find out the number of mole of acid (or alkali) left after a main
reaction.
After finding the number of mole of acid (or alkali), the number of mole of acid (or alkali) reacted
in the main reactioncan be calculated.
e.g. An excess amount of acid (or alkali) are added to react with X
Main reaction:
X + Acid (or alkali) →Product
(Excess)
The amount of acid (or alkali) left can be found by reacting with a standard alkali (or acid) solution.
Tiitrationsolution:
Standard alkali + Acid (Left) →Product
or Standard acid + Alkali (Left) →Product

24. Back Titration
Back titration
Calculation procedure:
1. Calculate the number of mole of standard alkali (or acid) used.
2. Calculate the number of mole of acid (or alkali) left.
3. Calculate the number of mole of acid (or alkali) reacted with X in the
main reaction by
4. Calculate the number of mole of X.
Number of mole of
acid (or alkali) reacted
with X
Number of mole of
acid (or alkali) added
at the start
Number of mole of
acid (or alkali) left
= -

25. Calculation Problems on Back Titration
EXERCISE:
2. 25.0 cm
3
of a 1.0 M solution of sodium hydroxide were placed in a flask. 1.40 g of an impure specimen of ammonium
chloride was added.
The flask and its contents were then carefully heated until no more ammonia gas was evolved.
The resulting solution was found to be alkaline and was diluted to exactly 250 cm
3
. 50 cm
3
of this solution required 5.10 cm
3
of 0.1 M hydrochloric acid for neutralization.
Calculate the percentage purity of the original impure ammonium chloride.
Answer:
NaOH(aq) + NH
4Cl(aq) → NH
3(g) + NaCl(aq) + H
2O(l)
NaOH(aq) + HCl(aq) → NaCl(aq) + H
2O(l)
no. of moles of HCl = (0.1M)(0.0051 dm
3
) = 5.1×10
-4
mol
no. of moles of NaOH in the 50 cm
3
solution = 5.1×10
-4
mol
no. of moles of NaOH in the 250 cm
3
solution = (5.1×10
-4
mol)(250 / 50) = 2.55×10
-3
mol
The initial no. of moles of NaOH = (1.0M)(0.025 dm
3
) = 0.025 mol
The no. of moles of NaOH reacted with NH
4Cl = 0.025 mol -2.55×10
-3
mol = 0.02245 mol
no. of moles of NH
4Cl reacted = 0.02245 mol
mass of NH
4Cl = (0.02245)(14.0 + 1.0×4 + 35.5) = 1.201 g
percentage purity = (1.201 / 1.4)(100%) = 85.79%

26. Calculation Problems on Back Titration
HOMEWORK :
1.Adrugtabletcanrelievestomachaches.Itcontainsaluminiumhydroxide(Al(OH)
3),astheonlyactive
ingredient.Astudentperformedthefollowingexperimenttodeterminetheamountofaluminiumhydroxide
containedinonedrugtablet.
Step1:Onedrugtabletwasdissolvedin50.0cm
3
of1.00moldm
-3
hydrochloricacidtoformasolution.
Step2:Thesolutionwasplacedinavolumetricflaskandthendilutedto250.0cm
3
withdistilledwater.
Step3:25.0cm
3
ofthedilutedsolutionweretitratedwith0.190moldm
-3
sodiumhydroxidesolutionusinga
suitableindicator.18.1cm
3
ofsodiumhydroxidesolutionwereneededtoreachtheendpoint.
(a)Calculatethenumberofmolesofexcesshydrochloricacidin25.0cm
3
ofthedilutedsolutionfrom
thedataobtainedinthetitration.
(b)CalculatethenumberofmolesofhydrochloricacidthatwasaddedinStep1.
(c)Hencecalculatethemassofaluminiumhydroxideinonedrugtablet.
(Relativeatomicmasses:H=1.0,O=16.0,Al=27.0)

27. Calculation Problems on Back Titration
HOMEWORK :
2. The following experiment was carried out to determine the concentration of copper(II) ions in a copper(II)
nitrate solution:
Step 1: 25.0 cm
3
of 0.660 mol dm
-3
sodium hydroxide solution were added to 25.0 cm
3
of copper(II) nitrate
solution to precipitate out the copper(II) hydroxide.
Step 2: The copper(II) hydroxide was removed from the reaction mixture.
Step 3: The alkali left over in the reaction mixture was titrated against 0.200 mol dm
-3
hydrochloric acid using a
suitable indicator. 20.5 cm
3
of the acid were required to reach the end point.
(a) Write an ionic equation for the reaction that occurred in step 1.
(b) Suggest a method to remove the copper(II) hydroxide from the reaction mixture in step 2.
(c)(i) Based on the titration result in Step 3, calculate the number of moles of hydroxide ions left over in the
reaction mixture.
(ii) Calculate the number of moles of sodium hydroxide that was added in Step 1.
(iii) Calculate the concentration of copper(II) ions in the copper(II) nitrate solution.

28. Calculation Problems on Back Titration
HOMEWORK :

29. Calculations on Volumetric Analysis
EXERCISE:
1. 1.5 g of a dibasic acid containing two molecules of water of crystallization are dissolved in distilled water
and the solution made up to 250 cm
3
. 25 cm
3
of this solution require for neutralization 26.2 cm
3
of 0.1 M
sodium hydroxide solution.
Calculate the relative molecular mass of the crystalline solid acid.
Answer: Let the formula of the crystalline solid acid be H
2A.2H
2O.
H
2A(aq) + 2NaOH(aq) → Na
2A(aq) + 2H
2O(l)
no. of moles of the NaOH solution = (0.1M)(0.0262 dm3) = 0.00262 mol
no. of moles of H
2A in the 25 cm3 solution = (0.00262 mol) /2 = 0.00131 mol
no. of moles of H
2A in the 250 cm3 solution = (0.00131 mol)(250 / 25) = 0.0131 mol
Because 1 unit of H
2A.2H
2O(s) gives one unit of H
2A(aq)
therefore, the no. of moles of H
2A.2H
2O = 0.0131 mol
Since no. of moles = mass / molar mass
therefore, the relative molecular mass of the solid acid = 1.5 / 0.0131 = 114.5 (no unit!)

30. Calculations on Volumetric Analysis
EXERCISE:
3. 25.0 cm
3
of 0.0527 M sodium carbonate solution were pipetted into a conical flask. Two drops of methyl
orange were added. Sulphuricacid was then added from a burette, until the solution just turned from yellow to
red. The titration was repeated 2 more times.
Calculate the molarity of the sulphuricacid.
Answer
Na
2CO
3(aq) + H
2SO
4(aq) → Na
2SO
4(aq) + CO
2(g) + H
2O(l)
Average volume of sulphuricacid used = (25.70 + 25.90)/2 = 25.80 cm
3
no. of moles of Na
2CO
3= (0.0527M)(0.025dm
3
) = 1.3175×10
-3
mol
no. of moles of H
2SO
4= 1.3175×10
-3
mol
concentration of H
2SO
4= (1.3175×10
-3
mol) / 0.0258 dm
3
= 0.051 M
Titration 1 2 3
Finalburettereading(cm3) 26.90 27.90 26.90
Initialburettereading(cm3) 0.30 2.20 1.00
Volumeofacidadded(cm3) 26.60 25.70 25.90

31. Calculations on Volumetric Analysis
EXERCISE:
4. 2.17 g of ethanedioicacid (COOH)2.2H2O were made up to 250 cm3 of solution. 25.0 cm3 of the solution
required 29.3 cm3 of 0.119 M sodium hydroxide solution for complete neutralization. Find the basicity of the
acid.
Answer
molecular mass of the solid acid = (12.0 + 16.0×2 + 1.0)2 + 18.0×2 = 126.0
no. of moles of the solid acid = (2.17)/(126) = 0.01722 mol
since 1 unit of (COOH)
2
.
2H
2O(s) gives 1 unit of (COOH)
2(aq)
therefore, the no. of moles of (COOH)
2(aq) in 250cm
3
solution = 0.01722 mol
no.ofmoles of (COOH)
2(aq) in 25 cm
3
solution = 0.01722mol / 10 = 0.001722 mol
no. of moles of NaOH = (0.119M)(0.0293 dm
3
) = 0.00349 mol
no. of moles of NaOH/ no. of moles of (COOH)
2(aq) = 0.00349 / 0.001722 = 2.024
therefore, the basicity of the acid is 2.
(H
2A + 2NaOH → Na
2A + 2H
2O)

32. Calculations on Volumetric Analysis
EXERCISE:
5. 2.00 g of a solid dibasic acid (H
2A) were dissolved in water and made up to 250 cm3. 25.0 cm3 of this
solution required 31.8 cm3 of 0.10 M sodium hydroxide solution for complete neutralization. Find the molar
mass of the acid.
Answer
H
2A(aq) + 2NaOH(aq) → Na
2A(aq) + 2H
2O(l)
no. of moles of NaOH = (0.10M)(0.0318dm
3
) = 0.00318 mol
no. of moles of H
2A in the 25 cm
3
solution = 0.00318 mol /2 = 0.00159 mol
no. of moles of H
2A in the 250 cm
3
solution = 0.00159 mol ×10 = 0.0159 mol
molar mass of H
2A = (2.00g)/(0.0159 mol) = 125.8 g mol
-1

33. Calculations on Volumetric Analysis
EXERCISE:
6.0.186 g of a sample of hydrated sodium carbonate Na2CO3.nH2O required 15.0 cm3 of 0.10M sulphuric
acid for complete neutralization. Find the number of molecules of water of crystallization in the sample of
sodium carbonate.
Answer
Na
2CO
3(aq) + H
2SO
4(aq) → Na
2SO
4(aq) + CO
2(g) + H
2O(l)
no. of moles of H
2SO
4= (0.10 M)(0.015 dm
3
) = 0.0015 mol
no. of moles of Na
2CO
3(aq) reacted = 0.0015 mol
since 1 unit of Na
2CO
3
.
nH
2O(s) gives 1 unit of Na
2CO
3(aq) when dissolves in water.
therefore, no. of moles of Na
2CO
3
.
nH
2O(s) = 0.0015 mol
formula mass of Na
2CO
3
.
nH
2O(s) = 23.0×2 + 12.0 + 16.0×3 + 18n = 106 + 18n
no. of moles of Na
2CO
3
.
nH
2O(s) = 0.186 / 106+18n = 0.0015
n = 1

34. Calculations on Volumetric Analysis
EXERCISE:
7. 0.509 g of an impure sample of anhydrous potassium carbonate (containing potassium chloride as impurity)
dissolves in water. The resulting solution required 35.0 cm3 of 0.20 M hydrochloric acid for complete
neutralization. Find the percentage purity of the anhydrous potassium carbonate sample.
Answer
K
2CO
3(aq) + 2HCl(aq) → 2KCl(aq) + CO
2(g) + H
2O(l)
no. of moles of HCl = (0.20 M)(0.035 dm
3
) = 0.007 mol
no. of moles of K
2CO
3reacted = (0.007 mol) /2 = 0.0035 mol
mass of K
2CO
3= (0.0035 mol)(39.0×2 + 12.0 + 16.0×3) = 0.483 g
percentage purity = (0.483 g / 0.509 g) ×100% = 94.89%