Potable water treatment
Potable water treatment

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Potable water treatment

4.7 Summary

Water in its 'natural' state supports a complex, yet fragile, ecosystem. The ability of natural watercourses to sustain aquatic life depends on a variety of physical, chemical and biological conditions. Biodegradable compounds, nutrients and dissolved oxygen must be available for the metabolic activities of the algae, fungi, bacteria and protozoa which are at the lowest level of the food chain. In addition, plant and animal growth cannot occur outside narrow ranges of temperature and pH. Suspended solids can restrict the necessary light penetration for photosynthesis. Stratification, both thermal and saline, can hinder the transport of necessary nutrients. Dissolved carbon dioxide, bicarbonates, carbonates, nitrates, phosphates and hardness salts must all be present in the right amounts for the successful functioning of the life of the river. Biological monitoring, e.g. BMWP scores, can be used as an indication of the state of health of a stretch of river.

Figure 18
Figure 18 Comparison of the distribution of fauna in the estuaries of the Tees and the Tay (the horizontal axis is the distance along the estuary towards the sea)

Self-assessment question

Which of the following statements is true?

A 'Sewage fungus' is a single species of fungus.

B Carbon dioxide and oxygen both react with water but it is only carbon dioxide which dissociates.

C Escherichia coli is a bacterium which is intestinal in origin and also pathogenic. This makes it suitable as an indicator organism for the presence of domestic sewage.

D With reference to the bicarbonate–carbonate equilibrium in a water which has a high pH, carbon dioxide present will remain as carbon dioxide.

E Permanent and temporary hardness both contribute to scaling, but for permanent hardness the scaling is due to decreased solubility, while for temporary hardness the scaling is due to the formation of new insoluble compounds.


E is true.

A 'Sewage fungus' is a mixture of different species of bacteria, fungi, algae and protozoa.

B Oxygen does not react with water but with the substances contained in the water.

C E. coli, although common in animals and humans is not pathogenic.

D High pH values correspond to low H+ or alkaline conditions. From Equation (1) (describing the bicarbonate–carbonate equilibrium), this will mean a decrease in hydrogen ions and thus the equilibrium moves to the right, increasing the concentration of carbonate.

Self-assessment question

With reference to the bicarbonate–carbonate equilibrium, explain why the pH of water often rises during photosynthesis by aquatic plants.


According to Equation (1), the removal of carbon dioxide by plants causes the reaction to shift to the left to compensate. The concentration of hydrogen ions then drops and so the pH rises.

Self-assessment question

A sample of sea water from a sector of the North Sea contains 35 g of salt per 1000 g of water. If the sea water is found to be 30% saturated with oxygen at 8°C and at 760 mm barometric pressure, what is the oxygen content in grams per m3?


For the sea water to be saturated, the dissolved oxygen concentration at 8°C

But the sample is only 30% saturated. Thus dissolved oxygen present

Self-assessment question

A benthic sample taken half a kilometre downstream of an effluent discharge pipe in a river showed the following biological species:

  • mayfly nymph (Baetidae)

  • snails from two different families

  • leeches from two different families

  • fly larvae (Chironomidae and Culicidae)

  • true worms from three different families.

Calculate the BMWP score and comment on your result.


Species Points scored
Mayfly nymph Baetidae 4
Snails 2 families 3 × 2 = 6
Leeches 2 families 3 × 2 = 6
Fly larvae Chironomidae 2
Culicidae 1
True worms 3 families 1 × 3 = 3
Total 22

The BMWP score is 22. The low score and the absence of clean water indicators signifies that the effluent discharge is having a detrimental effect on the river quality.

Self-assessment question

Using Figure 17(b) from the previous page, determine:

(a) the salinity for the Tees (in g per 1000 g of water) at 11 km from the mouth of the river at the surface and at a depth of 3 m;

(b) the salinity for the Thames (in g per 1000 g) at 36 km from the mouth of the river at the surface and at a depth of 3 m.


Look at the diagram for the River Tees and go to 11 km on the scale given. (You will need to use a ruler to determine the exact spot.) Then draw a vertical line until you reach the top of the figure showing salinity. The top depicts the salinity at the surface. The value is between 19 g and 25 g per 1000 g of water. You will have to use a ruler again to determine the exact value.

For the salinity at 3 m depth, draw a horizontal line at the 3 m mark, until it hits the vertical line that you first drew. The point of intersection between the two lines gives the salinity, at 3 m depth, 11 km from the mouth of the river.

You will have to repeat the exercise for part (b) for the River Thames using the bottom diagram but for a distance of 36 km from the mouth of the river.

(a) 19.6 g and 25.6 g per 1000 g of water, respectively.

(b) 20.2 g per 1000 g of water at surface and also at 3 m depth.


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