VOLUMETRIC ANALYSIS _ CHEMISTRY
NECTA 2012
1. You are provided with the following solutions;
TZ: Containing 3.5g of impure sulphuric acid in 500 cm3 of solution;
LO: Contain 4g of sodium hydroxide in 1000 cm3 of solution
Phenolphthalein and methyl orange indicators.
Questions
i/ What is the suitable indicator for the titration of the given solution? Give a reason for your answer
ii/ write a balanced chemical equation for the reaction between TZ and LO.
iii/ Why it is important to swirl or shake the contents of the flask during the addition of acid?
a) Titrate the acid (in a burette) against the base (in a conical flask) using two drops of your indicator and obtained three titre values.
b) 
i/ Cm3 of acid required Cm3 of base for complete reaction.
i/ Cm3 of acid required Cm3 of base for complete reaction.
ii/ Showing your procedure clearly, calculate the percentage purity of TZ.
SOLUTION
a) i/ The suitable indicator for the titration of the given solution is either of the two indicators, it can be either methyl orange (mo) or phenolphthalein (p. o. p) indicators because the titration involves strong acid against strong base.
ii/ H2SO4 (aq) + 2NaOH (aq) Na2SO4 (aq) + H2O (L)
iii/ It is important to swirl or shake the contents of the flask during the addition of the acid to allow reaction to take place properly or completely as well as allowing easy recognition of sharp end point during titration.
b) Table of results
Titration number
|
Pilot
|
1
|
2
|
3
|
Final burette reading (cm3)
|
25.90
|
25.10
|
25.40
|
25.00
|
Initial burette reading (cm3)
|
0.00
|
0.00
|
0.00
|
0.00
|
Volume used or titre value (cm3)
|
25.90
|
25.10
|
25.40
|
25.00
|
Average titre value = V1 + V2 + V3
3
Va = 25.10 + 25.40 + 25.00
3
Va = 25.16 cm3
c) i/ 25.16 cm3 of acid solution required 25.00 cm3 of base for complete reaction.
ii/ Calculate for the percentage purity of TZ
Given that
Concentration (g/dm3) = Conc in g/dm3
So, its’ Molarity (mol/dm3) = Conc in g/dm3
Molar mass g/mol
= 4g/dm3
40g/mol
Molarity of solution LO (base) = 0.1mol/dm3
Molarity of solution TZ (mol/dm3) can be obtained from the relation
Ma Va = na
Mb Vb nb
Where:
Ma = Molarity of Acid (TZ)
Va =Volume of Acid (TZ)
Mb =Molarity of Base (LO)
Vb=Volume of Base (LO)
Na =Number of Acid (TZ)
Nb =Number of Base (LO)
We have
Ma = ?
Va =25.16
Mb = 0.1
Vb=25
Nb =2
Na =1
H2SO4 (aq) + 2NaOH (aq) Na2SO4 (aq) + H2O (L)
Na= 1 Nb= 2
Ma Va = na
Mb Vb nb
Ma = Mb X Vb X na
Va X nb
Ma = 0.1 X 25 X 1
25.16 X 2
Ma = 0.05M
Molarity of solution TZ (acid) = 0.05M.
Then,
Calculate the percentage purity of TZ
Percentage purity = Concentration of pure X 100%
Concentration of impure
Therefore
Find the concentration of pure in solution TZ
Concentration of pure = Molarity X molar mass
Conc of pure = 0.05 X 98
Concentration of pure = 4.9g/dm3
Find the concentration of impure in solution TZ
Concentration of impure = Mass
Volume
Conc of impure = 3.5g (Volume = 500 = 0.5 dm3)
0.5 1000
Concentration of impure = 7g/dm3
Percentage purity = Concentration of pure X 100%
Concentration of impure
Percentage purity = 4.9g/dm3 X 100%
7g/dm3
Percentage purity = 70%
v Percentage purity of solution TZ = 70%
1. You are provided with the following solutions;
Solution AA Prepared by diluting 100cm3 of 1M Hydrochloric Acid
Solution BB is sodium hydroxide
Phenolphthalein indicator (POP)
Procedure:
Pipette 20 cm3 or 25cm3 of solution BB into a titration flask. Add two drops of POP indicator. Titrate solution BB against solution AA from the burette until a colour change is observed. Note the burette reading. Repeat the procedure to obtain three more readings. Record your results as shown bellow.
a) Table of results
i) Burette readings
Titration
|
Pilot
|
1
|
2
|
3
|
Final reading (cm3)
| ||||
Intial reading (cm3)
| ||||
Volume used (cm3)
|
ii) 
The volume of piprtte used was Cm3
The volume of piprtte used was Cm3
iii) 
The colour change at the end point was from to
The colour change at the end point was from to
iv) 
Cm3 of solution AA was required to neutralize Cm3of solution BB
Cm3 of solution AA was required to neutralize Cm3of solution BB
b) Write a balanced chemical equation for the neutralization of sodium hydroxide by hydrochloric acid
c) Calculate the:
(i) Molarity of solution AA
(ii) Concetration in mole/dm3 of solution BB
(iii) Concentration in g/dm3 of solution BB
SOLUTION
a) Table of results
i) Burette readings
Titration
|
Pilot
|
1
|
2
|
3
|
Final reading (cm3)
|
24.00
|
48.10
|
23.70
|
47.00
|
Intial reading (cm3)
|
0.00
|
24.00
|
0.00
|
23.70
|
Volume used (cm3)
|
24.00
|
24.10
|
23.70
|
23.30
|
Average volume of acid (AA) = V1 + V2 + V3
3
Va = 24.10 + 23.70 + 23.30
3
Va = 23.70 cm3
ii) 
The volume of pipette used was 25.00 Cm3
The volume of pipette used was 25.00 Cm3
v) The colour change at the end point was from Pink to Colourless
vi) 
25.00 Cm3 of solution AA was required to neutralize 23.70 Cm3of solution BB
25.00 Cm3 of solution AA was required to neutralize 23.70 Cm3of solution BB
b) Balanced chemical equation
HCl (aq) + NaOH (aq) NaCl (aq) + H2O (L)
c)
(i) Molarity of solution AA
From the formula
McVc = MdVd
Where
Mc = Molarity of conc. Acid (AA)
Vc = Volume of conc. Acid (AA)
Md = Molarity of diluted Acid (AA)
Vd = Volume of diluted or final volume of Acid AA
We have
Mc =1M
Vc =100cm3
Vd =1000cm3
Md= ?
McVc = MdVd
1Mx100cm3 = Md x 1000cm3
Md = 1M x 100cm3
1000cm3
= 0.1M
v Molarity of solution AA = 0.1M
(ii) Concentration in mole/dm3 of solution BB
From the relation
Ma Va = na
Mb Vb nb
Where:
Ma = Molarity of Acid (AA)
Va =Volume of Acid (AA)
Mb =Molarity of Base (BB)
Vb=Volume of Base (BB)
Na =Number of Acid (AA)
Nb =Number of Base (BB)
HCl (aq) + NaOH (aq) NaCl (aq) + H2O (L)
Na= 1 Nb= 1
We have
Ma = 0.1
Va =23.70
Mb = ?
Vb=25
Nb =1
Na =1
Mb = Ma X VaX nb
Vb X na
Mb = 0.1 X 23.70 X 1
25X 1
Mb = 0.0948 M = 0.1M
v Concentration of BB in mol/dm3 = 0.1M.
(iii) Concentration in g/dm3 of solution BB
From the relation
Molarity (mol/dm3) = Conc in g/dm3
Molar mass g/mol
0.1 = Conc in g/dm3
40
Concentration g/dm3 = Molarity X molar mass
Conc = 0.1 x 40
= 4 g/dm3
v Concentration in g/dm3 of solution BB = 4 g/dm3
1. You are provided with the following solutions;
Ø Solution E: Containing 3.65g of Hydrochloric Acid per dm3 of solution.
Ø Solution K: Containing 3.575g of pure hydrated sodium carbonate, Na2CO3.xH2O per 0.25dm3 of solution.
Ø Methyl orange indicator solution
Procedure:
Put the acid solution into the burette. Pipette 20 cm3 or 25cm3 of solution K into a titration flask. Add few drops of methyl orange indicator. Titrate this base solution against the acid solution until an end point is reached. Record your results as shown bellow and Repeat the titration.
a) Table of results
Titration
|
Pilot
|
1
|
2
|
3
|
Final reading (cm3)
| ||||
Intial reading (cm3)
| ||||
Volume used (cm3)
|
The volume of pipette used was Cm3
i/ The colour change at the end point was from to
a) 
i/ The volume of acid solution E needed for complete neutralization was Cm3
i/ The volume of acid solution E needed for complete neutralization was Cm3
ii/ Write down the balanced chemical equation for the reaction
b) Calculate the:
i/ Molarity of Acid solution E
ii/ Molarity of base solution K
iii/ Find the value of water crystallization in Na2CO3. XH2O
SOLUTION
The volume of pipette used was 20.00 Cm3
a) Table of results
Titration
|
Pilot
|
1
|
2
|
3
|
Final reading (cm3)
|
20.00
|
39.80
|
19.60
|
40.00
|
Intial reading (cm3)
|
0.00
|
20.00
|
0.00
|
20.00
|
Volume used (cm3)
|
20.00
|
19.80
|
19.60
|
20.00
|
Average volume of acid (E) = V1 + V2 + V3
3
Va = 19.80 + 19.60 + 20.00
3
Va = 19.93 cm3
b) i/ The colour change at the end point was from Yellow to pink
ii/ Balanced equation for the reaction
2HCl (aq) + Na2CO3(aq) 2NaCl (aq) + H2O (L) + CO2(g)
C) Calculate the;
(i) Molarity of Acid solution E
From the relation
Molarity (mol/dm3) = Conc in g/dm3
Molar mass g/mol
Then
Find the concentration of solution E
Concentration of solution E = Mass
Volume
Conc of E = 3.65g
1dm3
Conc of E= 3.65g/dm3
Therefore;
Molarity (mol/dm3) = Conc in g/dm3
Molar mass g/mol
Molarity = 3.65g/dm3
36.5g/mol
Molarity = 0.1 mol/dm3
v Molarity of Acid of solution E = 0.1 mol/dm3
(ii) Molarity of base solution K
From the relation
Ma Va = na
Mb Vb nb
Where:
Ma = Molarity of Acid (E)
Va =Volume of Acid (E)
Mb =Molarity of Base (K)
Vb=Volume of Base (K)
Na =Number of Acid (E)
Nb =Number of Base (K)
2HCl (aq) + Na2CO3(aq) 2NaCl (aq) + H2O (L) + CO2(g)
Na= 2 Nb= 1
We have
Ma = 0.1M
Va =19.93 cm3
Mb = ?
Vb=20.00 cm3
Nb =1
Na =2
Mb = Ma X VaX nb
Vb X na
Mb = 0.1 X 19.93 X 1
20X 2
Mb = 0.0498
Mb = 0.05 M
v Molarity of base (Mb) = 0.05M.
(iii) The value of water crystallization in Na2CO3. XH2O
Find the Concentration in g/dm3 of Na2CO3
From the formular
Molarity (mol/dm3) = Conc in g/dm3
Molar mass g/mol
Then
Conc. Of Na2CO3 = Molarity X Molar mass
Conc. Of Na2CO3 = 0.05M X 106g/mol
Concentration of solution K = 5.3g/dm3
Then
Find the molar mass of Na2CO3. XH2O
From the formula
Molarity (mol/dm3) = Conc in g/dm3
Molar mass g/mol
Then
Molar mass (g/mol) = Conc in g/dm3
Molarity (mol/dm3)
But
Concentration of solution K = Mass
Volume
Conc of k = 3.575g
0.25dm3
Conc of K= 14.3g/dm3
Also
Molar mass (g/mol) = Conc in g/dm3
Molarity (mol/dm3)
Molar mass = 14.3g/dm3
0.05mol/dm3
= 286g/mol
Molar mass of Na2CO3.xH2O is 286g/mol
Therefore,
Na2CO3.xH2O = 286
(23 X 2 +12+16X3 +(2+16) = 286
46 + 12 + 48 + 18x
106 + 18x = 286
18x = 286- 106
18x = 180
18x =180
18 18
X = 10
v The value of x in Na2CO3.xH2O is 10
TECNIC QUESTION
1. You are provided with the following solutions;
Solution EE: Containing 0.2 M of sodium hydroxide
Solution FF: Containing 4.50g of acid H2Q which were dissolved in 250 cm3 of solution
Phenolphtalein indicator and Methyl orange indicator
Procedure;
(i) Measure exactly 75 cm3 of acid by using measuring cylinder and then transfer it into an empty clean beaker.
(ii) Measure exactly 75 cm3 of distilled water and then put it into the solution of the acid in part (i) above.
(iii) Stir well the solution mixture in part (ii) above by means of a clean and dry glass rod
Questions
a. (i) Why is it important to clean a conical flask at the end of each titration before staring the subsequent titration?
(ii)State the role of white tile or white paper on which conical flask and its contents is put during titration.
b. Titrate solution EE against solution FF three times
c. 
Cm3 of sodium hydroxide required cm3 of acid H2Q for complete neutralization.
Cm3 of sodium hydroxide required cm3 of acid H2Q for complete neutralization.
d. Identify H2Q acid by providing its name and formula.
e. Classify acid H2Q according to its strength, basicity and origin respectively and sate why in each case.
f. i/ Was methyl orange also a suitable indicator for this titration? Give reasons.
ii/ What is the significance of the indicator in this experiment?
iii/ Why there a colour change when enough acid has been added into the base
SOLUTION
a. (i) It is important to clean a conical flask at the end of each titration before starting the subsequent titration in order to avoid the effect of impurities in the conical flask which will turn the effect the titre value the acid used. Or to avoid effect of diluting base solution in the next experiment due to waste
(ii)The role of white tile or white paper on which conical flask and its contents is put during titration was to increase clear visibility end point of colour.
b. Table of results of burette reading
Titration
|
Pilot
|
1
|
2
|
3
|
Final reading (cm3)
|
20.00
|
39.80
|
19.70
|
40.00
|
Intial reading (cm3)
|
0.00
|
20.00
|
0.00
|
20.00
|
Volume used (cm3)
|
20.00
|
19.80
|
19.00
|
20.00
|
Average volume of acid (E) = V1 + V2 + V3
3
Va = 19.80 + 19.70 + 20.00
3
Va = 19.97 cm3
c. (i) 20.00cm3 sodium hydroxide required 19.97cm3 of acid H2Q for complete neutralization
(ii)Colour change from Pink to colourless point of equivalent
d. Identification of acid H2Q and name it’s the formula
Step 1: calclat molarity of diluted acid H2Q.
From the relation
Ma Va = na
Mb Vb nb
Where:
Ma = Molarity of Acid (FF)
Va =Volume of Acid (FF)
Mb =Molarity of Base (EE)
Vb=Volume of Base (EE)
Na =Number of Acid (FF)
Nb =Number of Base (EE)
H2Q (aq) +2 NaOH(aq) Na2Q (aq) + 2H2O (L) + CO(g)
Na= 1 Nb= 2
We have
Ma = ?
Va =19.97 cm3
Mb = 0.2 M
Vb=20.00 cm3
Nb =2
Na =1
Ma= Mb X VbX na
Va X nb
Ma = 0.2 X 20.00 X 1
19.97X 2
Ma = 0.1 M
Ma= 0.1 M
v Molarity of diluted acid = 0.1M.
But from
Find the molarity before diluting acid
From the formular
McVc = MdVd
Where
Mc = Molarity of conc. Acid (FF)
Vc = Volume of conc. Acid (FF)
Md = Molarity of diluted Acid (FF)
Vd = Volume of diluted or final volume of Acid FF
We have
Mc =?
Vc =75cm3
Vd =150cm3
Md=0.1
McVc = MdVd
MC x75cm3 = 0.1 x 150cm3
Mc = 0.1M x 150cm3
75cm3
= 0.2M
Molarity of conc of solution FF = 0.2M
Given that
4.50g = 250cm3
? = 1000
Concentration = 18g/dm3
Step 2:
Find the molar mass of H2Q
Molar mass (g/mol) = Conc in g/dm3
Molarity (mol/dm3)
Molar mass = 18g/dm3
0.2mol/dm3
Molar mass = 90g/mol
So,
H2Q =90
2 + Q=90
Q=90-2
Q= 88
Q= 88 (This is molecular mass of Q)
v Acid H2Q is most probably oxalic acid with the formula H2C2O4
e. Acid H2Q which is H2C2O4 is weak acid, because its degree of dissociation is low. For example dissociate partially compared to strong acids like hydrochloric acid which dissociate fully.
· Basically of H2Q is two (2) because it contain two (2) number of hydrogen atoms
· Original of oxalic acid or ethanedioic is organic acid or organic compound with structure
C O
OH
C O
OH
f. i/ Methyl orange indicators are not suitable indicator because the titration involves weak acid against strong base and if applied the colour will change before equivalent or end point.
ii/ The significance of the indicator in this experiment was to show the end point of the reaction. The indicator changes colour when the end point reached. The colour changes from Yellow to pink
iii/ When enough acid has been added into the base the reaction will complete or will neutralize base and reaction take place properly and the colour change to show end point of titration.
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