Worked Solutions
Chapter 11
Question 5
What is the concentration in grams per litre of (a) a 0.1 M HCl solution (b) a 0.05 M Na2CO3
solution (c) a 0.05 M NaOH solution (d) a 0.01 M KOH solution (e) a 0.2 M HNO3 solution?
Answer:
(a) Molar mass of HCl = 36.5 g mol-1
0.1M HCl solution contains 0.1 moles = 0.1 X 36.5 g in 1 litre
Concentration of HCl solution = 3.65 g l–1
(b) Molar mass of Na2CO3 = 106 g mol-1
0.05M Na2CO3 solution contains 0.05 moles = 0.05 X 106 g in 1 litre
Concentration of Na2CO3 solution = 5.3 g l–1
(c) Molar mass of NaOH = 40 g mol-1
0.05M NaOH solution contains 0.05 moles = 0.05 X 40 g in 1 litre
Concentration of NaOH solution = 2 g l–1
(d) Molar mass of KOH = 56 g mol-1
0.01M KOH solution contains 0.01 moles = 0.01 X 56 g in 1 litre
Concentration of KOH solution = 0.56 g l–1
(e) Molar mass of HNO3 = 63 g mol-1
0.2M HNO3 solution contains 0.2 moles = 0.2 X 63 g in 1 litre
Concentration of HNO3 solution = 12.6 g l–1
Question 6
A solution contains 25 g of potassium hydroxide (KOH) in 1 litre of solution. Calculate the
concentration of the solution in moles per litre.
Answer:
25 g KOH = 25/56 moles KOH = 0.45 moles KOH
Concentration of the solution = 0.45 mol l-1 (=0.45 M)
Question 7
The concentration of a potassium manganate(VII) solution (KMnO4) is 6.32 g l –1. What is the
concentration of this solution in moles per litre?
Answer:
6.32 g KMnO4 = 6.32/158 moles KMnO4 = 0.04 moles KMnO4
Concentration of the solution = 0.04 mol l-1 (= 0.04 M)
1
Question 8
A solution contains 4 g of NaOH in 250 cm3 of solution. Calculate the concentration of the
solution in moles per litre.
Answer:
4 g of NaOH in 250 cm3 of solution is equivalent to 4 x 1000/250 = 16 g in 1 litre
16 g NaOH = 16/40 moles NaOH = 0.4 moles NaOH
Concentration of the solution = 0.4 mol l-1 (= 0.4 M)
Question 9
A solution contains 4.9 g H2SO4 in 100 cm3 of solution. Calculate the concentration of the
solution in moles per litre.
Answer:
4.9 g of H2SO4 in 100 cm3 of solution is equivalent to 4.9 x 1000/100 = 49 g in 1 litre
49 g H2SO4 = 49/98 moles H2SO4 = 0.5 moles H2SO4
Concentration of the solution = 0.5 mol l-1 (= 0.5 M)
Question 10
A bottle of wine contains 49 g of ethanol (C2H5OH) in 700 cm3 of solution. Calculate the
concentration of ethanol in moles per litre.
Answer:
49 g of C2H5OH in 700 cm3 of solution is equivalent to 49 x 1000/700 = 70 g in 1 litre
70 g C2H5OH = 70/46 moles C2H5OH = 1.52 moles C2H5OH
Concentration of the solution = 1.52 mol l-1 (= 1.52 M)
Question 11
What mass of potassium hydroxide (KOH) is contained in 500 cm3 of 0.05 M potassium
hydroxide solution?
Answer:
1,000 cm3 of 0.05 M KOH solution contains 0.05 moles
500 cm3 of 0.05 M KOH solution contains 0.05 X 500 / 1,000 moles
= 0.025 moles
Molar mass of KOH = 56 g mol-1
Mass of KOH contained in 500 cm3 solution = 56 X 0.025 g
= 1.4 g
2
Question 12
What mass of sodium hydroxide (NaOH) is contained in 250 cm3 of 0.02 M sodium
hydroxide solution?
Answer:
1,000 cm3 of 0.02 M NaOH solution contains 0.02 moles
250 cm3 of 0.02 M NaOH solution contains 0.02 X 250 / 1,000 moles
= 0.005 moles
Molar mass of NaOH = 40 g mol-1
Mass of NaOH contained in 250 cm3 solution = 40 X 0.005 g
= 0.2 g
Question 13
How many moles of hydrochloric acid (HCl) are contained in 25 cm3 of 0.05 M hydrochloric
acid solution?
Answer:
Number of moles = volume in cm3 X concentration in moles per litre / 1,000
= 25 X 0.05 / 1,000
= 0.00125
Question 14
How many moles of sulfuric acid (H2SO4) are contained in 15 cm3 of 0.02 M sulfuric acid
solution?
Answer:
Number of moles = volume in cm3 X concentration in moles per litre / 1,000
= 15 X 0.02 / 1,000
= 0.0003
Question 15
How many moles of sodium carbonate are there in 100 cm3 of 0.05 M sodium carbonate
solution?
Answer:
Number of moles = volume in cm3 X concentration in moles per litre / 1,000
= 100 X 0.05 / 1,000
= 0.005
Question 16
The concentration of dissolved oxygen in a water sample is 0.00002 M. Calculate the
concentration of dissolved oxygen in the water sample in p.p.m.
3
Answer:
Molar mass of O2 = 32 g mol-1
There are 0.00002 moles of dissolved oxygen in 1 litre of the water sample.
Mass of O2 in 1 litre = 0.00002 x 32 g = 0.00064 g = 0.00064 x 1,000 mg = 0.64 mg
Concentration of dissolved oxygen in the water sample = 0.64 p.p.m.
Question 17
A solution contains 25 g sodium hydroxide in 1 litre of solution. Express the concentration of
the solution in % (w/v).
Answer:
The solution contains 25 g NaOH in 1 litre.
100 cm3 of solution contains 25 x 100 / 1,000 g NaOH = 2.5 g NaOH
Concentration of NaOH in the solution = 2.5% (w/v)
Question 18
A solution contains 8 g sodium hydroxide in 500 cm3 of solution. Express the concentration
of the solution in % (w/v).
Answer:
The solution contains 8 g sodium hydroxide in 500 cm3 of solution.
100 cm3 of solution contains 8 x 500 / 1,000 g NaOH = 1.6 g NaOH
Concentration of NaOH in the solution = 1.6% (w/v)
Question 19
A bottle of vinegar contains 77 cm3 ethanoic acid in 700 cm3 of solution. Express the
concentration of ethanoic acid in the solution in % (v/v).
Answer:
The solution contains 77 cm3 ethanoic acid in 700 cm3 of solution.
100 cm3 of solution contains 77 x 100 / 700 cm3 ethanooic acid = 11 cm3 ethanoic acid
Concentration of ethanoic acid in the solution = 11% (v/v)
Question 20
A solution contains 20 g potassium hydroxide in 250 g of solution. Express the concentration
of the solution in % (w/w).
Answer:
The solution contains 20 g KOH in 250 g of solution.
100 g of solution contains 20 x 100 / 250 g KOH = 8 g KOH
Concentration of KOH in the solution = 8% (w/w)
4
Question 21
The label on a bottle of wine indicates that the concentration of alcohol in the wine is
9% (v/v). What volume of alcohol is there in 250 cm3 of the wine?
Answer:
100 cm3 of the wine contains 9 cm3 alcohol.
250 cm3 of the wine contains 9 x 250 / 100 cm3 alcohol = 22.5 cm3 alcohol.
Question 22
What volume of ethanol is there in 500 cm3 of a 5% solution of ethanol in water?
Answer:
100 cm3 of the solution contains 5 cm3 ethanol.
500 cm3 of the solution contains 5 X 500 / 100 cm3 ethanol = 25 cm3 ethanol.
Question 26
If 25 cm3 of a 1 M hydrochloric acid solution is diluted to a volume of 500 cm3 with water,
what is the concentration of the diluted solution?
Answer:
Vdil X Mdil = Vconc X Mconc
500 X Mdil = 25 X 1
Mdil = 25 X 1 / 500 mol l-1 = 0.05 mol l-1
Concentration of the diluted solution = 0.05 mol l-1
Question 27
If 10 cm3 of a 0.1 M potassium hydroxide solution is diluted to a volume of 250 cm3 with
water, what is the concentration of the diluted solution?
Answer:
Vdil X Mdil = Vconc X Mconc
250 X Mdil = 10 X 0.1
Mdil = 10 X 0.1 / 250 mol l-1 = 0.004 mol l-1
Concentration of the diluted solution = 0.004 mol l-1
Question 28
What volume of a 1 M hydrochloric acid solution is needed to make up 500 cm3 of a 0.1 M
hydrochloric acid solution?
Answer:
Vdil X Mdil = Vconc X Mconc
500 X 0.1 = Vconc X 1
5
Vconc = 500 X 0.1 / 1 mol l-1 = 50 cm3
Volume of 1 M hydrochloric acid solution needed = 50 cm3
Question 29
What volume of a 0.5 M sulfuric acid solution is needed to make up 250 cm3 of a 0.1 M
sulfuric acid solution?
Answer:
Vdil X Mdil = Vconc X Mconc
250 X 0.1 = Vconc X 0.5
Vconc = 250 X 0.1 / 0.5 mol l-1 = 50 cm3
Volume of 0.5 M sulfuric acid solution needed = 50 cm3
Question 40
In a titration, 25 cm3 of a 0.05 M sodium carbonate solution required 22 cm3 of a hydrochloric
acid solution for complete neutralisation. Calculate the concentration of the hydrochloric acid
solution.
The equation for the reaction is
2HCl(aq) + Na2CO3(aq)  2NaCl(aq) + H2O(l) + CO2(g)
Answer:
V1 x M1 x n2 = V2 x M2 x n1
22 x M1 x 1 = 25 x 0.05 x 2
M1 = 25 x 0.05 x 2 / (22 x 1)
= 0.11 mol l-1
Concentration of hydrochloric acid solution = 0.11 mol l-1
Question 41
In a titration, 25 cm3 of a sodium hydroxide solution required 19.5 cm3 of a 0.1 M
hydrochloric acid solution for complete neutralisation. Calculate the concentration of the
sodium hydroxide solution. The equation for the reaction is
HCl(aq) + NaOH(aq) 
NaCl(aq) + H2O(l)
Answer:
V1 x M1 x n2 = V2 x M2 x n1
19.5 x 0.1 x 1 = 25 x M2 x 1
M2 = 19.5 x 0.1 x 1 / (25 x 1)
= 0.078 mol l-1
Concentration of sodium hydroxide solution = 0.078 mol l-1
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Question 42
In a titration, 25 cm3 of a potassium hydroxide solution required 22.5 cm3 of a 0.01 M
hydrochloric acid solution for complete neutralisation. Calculate the concentration of the
potassium hydroxide solution. The equation for the reaction is
HCl(aq) + KOH(aq) 
KCl(aq) + H2O(l)
Answer:
V1 x M1 x n2 = V2 x M2 x n1
22.5 x 0.01 x 1 = 25 x M2 x 1
M2 = 22.5 x 0.01 x 1 / (25 x 1)
= 0.009 mol l-1
Concentration of potassium hydroxide solution = 0.009 mol l-1
Question 43
In a titration, what volume of a 0.1 M hydrochloric acid is required to neutralise 25 cm3 of a
0.02 M sodium carbonate solution? The equation for the reaction is
2HCl(aq) + Na2CO3(aq)  2NaCl(aq) + H2O(l) + CO2(g)
Answer:
V1 x M1 x n2 = V2 x M2 x n1
V1 x 0.1 x 1 = 25 x 0.02 x 2
V1 = 25 x 0.02 x 2 / (0.1 x 1)
= 10 cm3 of hydrochloric acid
Question 44
In a titration, what volume of a 0.05 M hydrochloric acid is required to neutralise 25 cm3 of a
0.04 M sodium hydroxide solution? The equation for the reaction is
HCl(aq) + NaOH(aq) 
NaCl(aq) + H2O(l)
Answer:
V1 x M1 x n2 = V2 x M2 x n1
V1 x 0.05 x 1 = 25 x 0.04 x 1
V1 = 25 x 0.04 x 1 / (0.05 x 1)
= 20 cm3 of hydrochloric acid
Question 45
What volume of 0.02 M H2SO4 is needed to neutralise 25 cm3 of a 0.05 M NaOH solution?
The equation for the reaction is
H2SO4(aq) + 2NaOH(aq) 
Na2SO4(aq) + 2H2O(l)
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Answer:
V1 x M1 x n2 = V2 x M2 x n1
V1 x 0.02 x 2 = 25 x 0.05 x 1
V1 = 25 x 0.05 x 1 / (0.02 x 2)
= 31.25 cm3 of sulfuric acid
Question 46
In a titration, 25 cm3 of a 0.12 M sodium hydroxide solution required 24 cm3 of a sulfuric acid
solution for complete neutralisation. Calculate (a) the number of moles of sodium hydroxide
consumed (b) the number of moles of sulfuric acid consumed and (c) the concentration of the
sulfuric acid solution. The equation for the reaction is
H2SO4(aq) + 2NaOH(aq) 
Na2SO4(aq) + 2H2O(l)
Answer:
(a) In the titration, 25 cm3 of 0.12 M sodium hydroxide solution were used.
1,000 cm3 of 0.12 M sodium hydroxide solution contains 0.12 moles.
25 cm3 of 0.12 M sodium hydroxide solution contains 0.12 X 25 / 1,000 moles
= 0.003 moles
(b) The equation for the reaction is
H2SO4(aq) + 2NaOH(aq)  Na2SO4(aq) + 2H2O(l)
2 moles of NaOH react with 1 mole of H2SO4
1 mole of NaOH reacts with 0.5 moles of H2SO4
0.003 moles of NaOH reacts with 0.003 X 0.5 moles of H2SO4
= 0.0015 moles of H2SO4
(c) In 24 cm3 of the H2SO4 solution there are 0.0015 moles of H2SO4
In 1,000 cm3 of the H2SO4 solution there are
0.0015 X 1,000 / 24 moles = 0.0625 moles
Concentration of the sulfuric acid solution = 0.0625 mol l-1
Question 47
20 cm3 of a 0.125 M HCl solution were neutralised by 25 cm3 of a KOH solution. The
equation for the reaction is
HCl(aq) + KOH(aq) 
KCl(aq) + H2O(l)
(a) How many moles of HCl were consumed?
(b) How many moles of KOH were neutralised?
(c) How many moles of KOH were there in 25 cm3 of the KOH solution?
(d) Calculate the concentration in mol l-1 of the KOH solution.
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Answer:
(a) In the titration, 20 cm3 of 0.125 M HCl solution were used.
1,000 cm3 of 0.125 M HCl solution contains 0.125 moles.
20 cm3 of 0.125 M HCl solution contains 0.125 X 20 / 1,000 moles
= 0.0025 moles
(b) The equation for the reaction is
HCl(aq) + KOH(aq) 
KCl(aq) + H2O(l)
One mole of HCl reacts with one mole of KOH.
0.0025 moles of HCl react with 0.0025 moles of KOH.
(c) The 25 cm3 of KOH consumed contained 0.0025 moles.
(d) In 25 cm3 of the KOH solution there are 0.0025 moles of KOH
In 1,000 cm3 of the KOH solution there are
0.0025 X 1,000 / 25 moles = 0.1 moles
Concentration of the KOH solution = 0.1 mol l-1
Question 48
What mass of magnesium will react with 20 cm3 of a 0.09 M hydrochloric acid solution? The
equation for the reaction is
2HCl(aq) + Mg(s) 
MgCl2(aq) + H2(g)
Answer:
1,000 cm3 of 0.09 M HCl solution contains 0.09 moles
20 cm3 of 0.09 M HCl solution contains 0.09 X 20 / 1,000 moles
= 0.0018 moles
From the equation, 1 mole of magnesium reacts with 2 mole of hydrochloric acid
0.0018 moles of hydrochloric acid react with 0.0018 / 2 = 0.0009 moles of magnesium
Molar mass of magnesium = 24 g mol-1
0.0009 moles of magnesium = 0.0009 X 24 g = 0.0216 g magnesium
Mass of magnesium = 0.022 g
Question 49
What mass of magnesium will react with 15 cm3 of a 0.12 M sulfuric acid solution? The
equation for the reaction is
H2SO4(aq) + Mg (s) 
MgSO4(aq) +
H2(g)
Answer:
1,000 cm3 of 0.12 M H2SO4 solution contains 0.12 moles
15 cm3 of 0.12 M H2SO4 solution contains 0.12 X 15 / 1,000 moles
9
= 0.0018 moles
From the equation, 1 mole of magnesium reacts with 1 mole of sulfuric acid
0.0018 moles of magnesium react with 0.0018 moles of sulfuric acid
Molar mass of magnesium = 24 g mol-1
0.0018 moles of magnesium = 0.0018 X 24 g = 0.0432 g magnesium
Mass of magnesium = 0.043 g
Question 50
A sample of vinegar was diluted from 25 cm3 to 250 cm3 with water. In a titration, 25 cm3 of a
0.1 M sodium hydroxide solution required 30 cm3 of the diluted vinegar for complete
neutralisation. Calculate the concentration of ethanoic acid in the vinegar in (a) mol l-1 (b) %
(w/v). The equation for the reaction is
CH3COOH(aq) + NaOH(aq) 
CH3COONa(aq) + H2O(l)
Answer:
(a) The formula V1 x M1 x n2 = V2 x M2 x n1 is used first to find the concentration of ethanoic
acid in the diluted vinegar.
V1 x M1 x n2 = V2 x M2 x n1
30 x M1 x 1 = 25 x 0.1 x 1
M1 = 25 x 0.1 x 1 / (30 x 1)
= 0.083 mol l-1
Concentration of ethanoic acid in the diluted vinegar = 0.083 mol l-1
The vinegar was originally diluted from 25 cm3 to 250 cm3 with water, i.e. a tenfold dilution.
Concentration of ethanoic acid in the vinegar = 10 x 0.083 mol l-1 = 0.83 M
(b) Molar mass of ethanoic acid (CH3COOH) = 60 g mol-1
Concentration of ethanoic acid in g l-1 in vinegar = 0.83 x 60 = 49.8
Percentage (w/v) of ethanoic acid in vinegar = 49.8 x 100 / 1000
= 4.98%
Question 51
2.91 g of a monobasic acid, HX, were dissolved in water and made up to 250 cm3 with water.
This solution was titrated with 0.108 M sodium hydroxide solution. 25 cm3 of the sodium
hydroxide solution required 22.5 cm3 of the HX solution for complete neutralisation. The
equation for the reaction is
HX (aq) + NaOH(aq)  NaX(aq) + H2O(l)
(a) Calculate the concentration in (i) g l-1 (ii) mol l-1 of the acid.
(b) Calculate the molar mass of HX.
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Answer:
(a) (i) 2.91 g of HX in 250 cm3 is equivalent to 2.91 X 4 g = 11.64 g in 1 litre. Thus the
concentration is 11.64 g l-1
(ii)
V1 x M1 x n2 = V2 x M2 x n1
22.5 x M1 x 1 = 25 x 0.108 x 1
M1 = 25 x 0.108 x 1 / (22.5 x 1)
= 0.12 mol l-1
Concentration of HX solution = 0.12 mol l-1
(b) 0.12 moles has a mass of 11.64 g
1 mole has a mass of 11.64/0.12 g = 97 g
The molar mass of HX is 97 g mol-1
Question 52
Crystals of hydrated sodium carbonate (Na2CO3.xH2O) of mass 3.15 g were dissolved in
water and made up to 250 cm3 in a volumetric flask. 25 cm3 of this solution required 15 cm3
of a 0.15 M hydrochloric acid solution for complete neutralisation. The equation for the
reaction is
2HCl (aq) + Na2CO3 (aq)  2NaCl (aq) + H2O(l) + CO2(g)
Find (a) the concentration of the sodium carbonate solution (b) the value of x in the formula
Na2CO3.xH2O (c) the percentage of water of crystallisation in the hydrated sodium carbonate.
Answer:
(a)
V1 x M1 x n2 =
V2 x M2 x n1
15 x 0.15 x 1 =
25 x M2 x 2
M2 =
15 x 0.15 x 1 / (25 x 2)
= 0.045 mol l-1
Concentration of the sodium carbonate solution = 0.045 mol l-1
(b) Molar mass of Na2CO3.xH2O = 46 + 12 + 48 + 18x = 106 + 18x
Concentration of sodium carbonate in mol l-1
= 0.045
Concentration of sodium carbonate in g l-1 = 3.15 X 4 = 12.6
Mass of 0.045 moles of Na2CO3.xH2O
Molar mass of Na2CO3.xH2O
= 12.6 g
= 12.6 / 0.045 g mol-1
= 280 g mol-1
106 + 18x = 280
18x = 174
x = 9.7
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(c) Molar mass of hydrated sodium carbonate = 280 g mol-1
Percentage of water of crystallisation in the compound = 174 x 100 / 280% = 62.14%
Question 53
In a titration, 25 cm3 of a potassium hydroxide solution required 27 cm3 of a 0.12 M sulfuric
acid solution for complete neutralisation. Calculate the concentration of the potassium
hydroxide solution in (a) mol l-1 (b) g l-1. The equation for the reaction is
H2SO4(aq) + 2KOH(aq) 
K2SO4(aq) + 2H2O(l)
Answer:
(a)
V1 x M1 x n2
= V2 x M2 x n1
27 x 0.12 x 2
= 25 x M2 x 1
= 27 x 0.12 x 2 / (25 x 1)
M2
= 0.2592 mol l-1
Concentration of potassium hydroxide solution = 0.26 mol l-1
(b) Molar mass of potassium hydroxide = 56 g mol-1
Concentration of KOH = 0.26 x 56 g l-1 = 14.6 g l-1
Question 54
In a titration, 20 cm3 of 0.1 M potassium hydroxide solution required 25 cm3 of a hydrochloric
acid solution for complete neutralisation. Calculate the concentration of the hydrochloric acid
solution in (a) mol l-1 (b) g l-1. The equation for the reaction is
HCl(aq) + KOH(aq) 
KCl(aq) + H2O(l)
Answer:
(a)
V1 x M1 x n2 = V2 x M2 x n1
25 x M1 x 1 = 20 x 0.1 x 1
M1 = 20 x 0.1 x 1 / (25 x 1)
= 0.08 mol l-1
Concentration of hydrochloric acid solution = 0.08 mol l-1
(b) Molar mass of hydrochloric acid = 36.5 g mol-1
Concentration of HCl = 0.08 X 36.5 g l-1 = 2.92 g l-1
Question 56
In a titration, 25 cm3 of a sodium carbonate solution required 22 cm3 of a 0.1 M hydrochloric
acid solution for complete neutralisation. Calculate the concentration of the sodium carbonate
solution in (a) mol l-1 (b) g l-1. The equation for the reaction is:
2HCl(aq) + Na2CO3(aq)  2NaCl(aq) + H2O(l)
+ CO2(g)
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Answer:
(a)
V1 x M1 x n2 = V2 x M2 x n1
22 x 0.1 x 1 = 25 x M2 x 2
M2 = 22 x 0.1 x 1 / (25 x 2)
= 0.044 mol l-1
Concentration of sodium carbonate solution = 0.044 mol l-1
(b) Molar mass of sodium carbonate = 106 g mol-1
Concentration of sodium carbonate solution = 0.044 x 106 g l-1 = 4.66 g l-1
Question 57
What volume of 0.01 M sodium hydroxide solution is required to neutralise 20 cm3 of a
hydrochloric acid solution containing 0.73 g of hydrogen chloride per 250 cm3 of solution?
The equation for the reaction is
HCl(aq) + NaOH(aq)  NaCl(aq) + H2O(l)
Answer:
0.73 g of hydrogen chloride per 250 cm3 of solution is equivalent to 0.73 x 4 g l-1 = 2.92 g l-1
= 2.92 / 36.5 mol l-1 = 0.08 mol l-1
V1 x M1 x n2 = V2 x M2 x n1
20 x 0.08 x 1 = V2 x 0.01 x 1
V2 = 20 x 0.08 x 1 / (0.01 x 1)
= 160 cm3 of hydrochloric acid
Question 58
What volume of a 0.01 M sodium carbonate solution is required to neutralise 25 cm3 of a
solution containing 0.98 g l-1 of sulfuric acid? The equation for the reaction is
H2SO4(aq) + Na2CO3(aq)  Na2SO4(aq) + H2O(l) + CO2(g)
Answer:
Concentration of sulfuric acid solution = 0.98 g l-1 = 0.98 / 98 mol l-1 = 0.01 mol l-1
V1 x M1 x n2 = V2 x M2 x n1
25 x 0.01 x 1 = V2 x 0.01 x 1
V2 = 25 x 0.01 x 1 / (0.01 x 1)
= 25 cm3 of sodium carbonate solution
Question 59
In a titration, 25 cm3 of a 0.1 M sodium hydroxide solution required 28 cm3 of a sulfuric acid
solution for complete neutralisation. Calculate (a) the number of moles of sodium hydroxide
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consumed (b) the number of moles of sulfuric acid consumed and (c) the concentration of the
sulfuric acid solution in mol l-1. The equation for the reaction is
H2SO4(aq) + 2NaOH(aq) 
Na2SO4(aq) + 2H2O(l)
Answer:
(a) In the titration, 25 cm3 of 0.1 M NaOH solution were used.
1,000 cm3 of 0.1 M NaOH solution contains 0.1 moles.
25 cm3 of 0.1 M NaOH solution contains 0.1 X 25 / 1,000 moles
= 0.0025 moles
(b) The equation for the reaction is
H2SO4(aq) + 2NaOH(aq) 
Na2SO4(aq) + 2H2O(l)
One mole of H2SO4 reacts with two moles of NaOH.
0.0025/2 moles = 0.00125 moles of H2SO4 react with 0.0025 moles of NaOH.
The number of moles of sulfuric acid consumed is 0.00125 moles.
(c) In 28 cm3 of the H2SO4 solution there are 0.00125 moles of H2SO4
In 1,000 cm3 of the H2SO4 solution there are
0.00125 X 1,000 / 28 moles = 0.045 moles
Concentration of the H2SO4 solution = 0.045 mol l-1
Question 60
Sodium hydrogencarbonate solution reacts with hydrochloric acid solution according to the
equation
HCl(aq) + NaHCO3(aq)  NaCl(aq) + H2O(l) + CO2(g)
It was found that 25 cm3 of a solution containing 8.4 g l-1 reacted with 27 cm3 of a solution of
hydrochloric acid. Calculate the concentration of the hydrochloric acid solution in (a) mol l-1
(b) g l-1.
Answer:
(a)
Concentration of sodium hydrogencarbonate solution = 8.4 g l-1 = 8.4 / 84 mol l-1 = 0.1 mol l-1
V1 x M1 x n2 = V2 x M2 x n1
27 x M1 x 1 = 25 x 0.1 x 1
M1 = 25 x 0.1 x 1 / (27 x 1)
= 0.093 mol l-1
(b) Concentration of hydrochloric acid solution = 0.093 mol l-1 = 0.093 x 36.5 mol l-1
= 3.39 g l-1
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