Worked Solutions
Chapter 21
Question 16
From the graph, the concentration of chlorine in the sample is 15.7 p.p.m.
Question 21
In an experiment to measure the total suspended solids in a water sample, a dry filter paper
was found to have a mass of 1.35g. One litre of water was passed through the filter paper,
which was then dried slowly and reweighed. The mass of the filter paper was found to have
increased to 1.41g. Calculate the total suspended solids in p.p.m.
Answer:
Total suspended solids = 1.41 – 1.35 = 0.06 g in 1l
= 0.06 g l-1
= 0.06 x 1000 mg l-1
= 60 p.p.m.
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Question 24
When 800 cm3 of filtered water is evaporated to dryness, 0.075g of solid substances remain in
the beaker. Calculate the total dissolved solids in p.p.m.
Answer:
Total dissolved solids = 0.075 g in 800 cm3
= 0.075 x 1000
----------------- g l-1
800
= 0.0938 mg l-1
= 94 p.p.m.
Question 33
50 cm3 of a hard water sample required 10.2 cm3 of a 0.01 M edta solution for complete
complexing of the metal ions. Calculate the total hardness of the water in p.p.m. of calcium
carbonate.
Answer:
H2Y2- + M2+ → MY2- + 2H+
(the ratio of edta ion : metal ion = 1:1)
V1 x M1 x n2 = V2 x M2 x n1
50.0 x M1 x 1 = 10.2 x 0.01 x 1
M1 = 10.2 x 0.01 x 1 / (50.0 x 1)
= 0.00204 moles/litre of Ca2+ and Mg2+
= 0.00204 x 100 g/l CaCO3
= 0.204 g/l CaCO3
= 0.204 x 1000 p.p.m. CaCO3
= 204 p.p.m. CaCO3
= total hardness of water sample.
Question 36
100 cm3 of water was analysed for dissolved oxygen using the Winkler method. The iodine
liberated required 20.0 cm3 of a 0.005 M sodium thiosulfate solution for complete reaction.
Calculate the dissolved oxygen content of the water.
Answer:
V1 x M1 x n2 = V2 x M2 x n1
100.0 x M1 x 4 = 20.0 x 0.005 x 1
M1 = 20.0 x 0.005 x 1 / (100.0 x 4)
= 0.00025
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Concentration of dissolved oxygen = 0.00025 M
= 0.00025 x 32 g/l
= 0.008 g/l
= 8 p.p.m.
Question 40
(b) A 1 litre sample of water was filtered into a flask, and it was found that the mass of the
filter paper, on drying, had increased by 0.16 g. The water was the evaporated, and it was
found that the mass of the flask had increased by 0.42 g. Calculate (i) the total suspended
solids in p.p.m. (ii) the total dissolved solids in p.p.m. in the water sample.
Answer:
(i) Total
suspended solids = 0.16 g in 1 l
= 0.16 g l-1
= 0.16 x 1000 mg l-1
= 160 p.p.m.
(ii) Total dissolved solids = 0.42 g in 1 l
= 0.42 g l-1
= 420 mg l-1
= 420 p.p.m.
Question 41
150 cm3 of water was analysed for dissolved oxygen using the Winkler method. The iodine
liberated required 36.0 cm3 of a 0.005 M sodium thiosulfate solution for complete reaction.
Calculate the dissolved oxygen content of the water.
Answer:
V1 x M1 x n2 = V2 x M2 x n1
150.0 x M1 x 4 = 36.0 x 0.005 x 1
M1 = 36.0 x 0.005 x 1 / (150.0 x 4)
= 0.0003
Concentration of dissolved oxygen = 0.0003 M
= 0.0003 x 32 g l-1
= 0.0096 g l-1
= 9.6 p.p.m.
Question 42
In an experiment to determine the B.O.D. of a water sample, the following results were
obtained for 100 cm3 samples of the water:
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Titre of 0.025 M sodium thiosulfate solution prior to incubation = 5 cm3
Titre of 0.025 M sodium thiosulfate solution after incubation = 3 cm3.
Calculate (a) the dissolved oxygen content of the water prior to incubation (b) the dissolved
oxygen content of the water after incubation (c) the B.O.D. of the water sample.
Answer:
(a) V1 x M1 x n2 = V2 x M2 x n1
100.0 x M1 x 4 = 5 x 0.025 x 1
M1 = 5 x 0.025 x 1 / (100.0 x 4)
= 0.000313
Concentration of dissolved oxygen = 0.000313 M
= 0.000313 x 32 g l-1
= 0.01 g l-1
= 10 p.p.m.
(b) V1 x M1 x n2 = V2 x M2 x n1
100.0 x M1 x 4 = 3 x 0.025 x 1
M1 = 3 x 0.025 x 1 / (100.0 x 4)
= 0.000188
Concentration of dissolved oxygen = 0.000188 M
= 0.000188 x 32 g l-1
= 0.00602 g l-1
= 6 p.p.m.
(c) BOD = dissolved oxygen content before incubation - dissolved oxygen content after
incubation
= 10 p.p.m. - 6 p.p.m.
= 4 p.p.m.
Question 43
50 cm3 of a hard water sample required 13.6 cm3 of a 0.01 M edta solution for complete
complexing of the metal ions. Calculate the total hardness of the water in p.p.m. of calcium
carbonate. If 200 cm3 of the hard water is boiled, and 50 cm3 samples of it are then titrated
with 0.01 M edta solution, it is found that the average titre of edta is 5 cm3. What is (a) the
permanent hardness
(b) the temporary hardness of the water sample?
Answer:
V1 x M1 x n2 = V2 x M2 x n1
50.0 x M1 x 1 = 13.6 x 0.01 x 1
M1 = 13.6 x 0.01 x 1 / (50.0 x 1)
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= 0.00272 moles/litre of Ca2+ and Mg2+
= 0.00272 x 100 g/l CaCO3
= 0.272 g/l CaCO3
= 0.272 x 1000 p.p.m. CaCO3
= 272 p.p.m. CaCO3
= total hardness of water sample.
(a) Boiling removes the temporary hardness in the sample, leaving only permanent hardness.
V1 x M1 x n2 = V2 x M2 x n1
50.0 x M1 x 1 = 5 x 0.01 x 1
M1 = 5 x 0.01 x 1 / (50.0 x 1)
= 0.001 moles/litre of Ca2+ and Mg2+
= 0.001 x 100 g/l CaCO3
= 0.1 g/l CaCO3
= 0.1 x 1000 p.p.m. CaCO3
= 100 p.p.m. CaCO3
= permanent hardness of water sample.
(b) Temporary hardness = Total hardness – Permanent hardness
= 272 – 100 = 172 p.p.m.
Question 45
100 cm3 of water was analysed at 288 K for dissolved oxygen using the Winkler method. The
iodine liberated required 4.0 cm3 of a 0.005 M sodium thiosulfate solution for complete
reaction. Calculate the dissolved oxygen content of the water in p.p.m.
Answer:
(a) V1 x M1 x n2 = V2 x M2 x n1
100.0 x M1 x 4 = 4 x 0.005 x 1
M1 = 4 x 0.005 x 1 / (100.0 x 4)
= 0.00005
Concentration of dissolved oxygen = 0.00005 M
= 0.00005 x 32 g l-1
= 0.0016 g l-1
= 1.6 p.p.m.
Question 47
In a titration it was found that 50 cm3 of a water sample required 7.5 cm3 of 0.01 M edta for
complete reaction.
(a) Calculate the total hardness of the water
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Answer:
V1 x M1 x n2 = V2 x M2 x n1
50.0 x M1 x 1 = 7.5 x 0.01 x 1
M1 = 7.5 x 0.01 x 1 / (50.0 x 1)
= 0.0015 moles/litre of Ca2+ and Mg2+
= 0.0015 x 100 g/l CaCO3
= 0.15 g/l CaCO3
= 0.15 x 1000 p.p.m. CaCO3
= 150 p.p.m. CaCO3
= total hardness of water sample.
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