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I NEED ALL QUESTIONS TO BE ANSWERED Comprehensive problem Draw a milliequivalent

ID: 479516 • Letter: I

Question

I NEED ALL QUESTIONS TO BE ANSWERED
Comprehensive problem
Draw a milliequivalents-per-liter diagram and list the hypothetical combinations for the following
groundwater data:
Calcium hardness = 175 mg/L Magnesium hardness = 40 mg/L
Sodium ion = 14 mg/L Potassium ion = 4 mg/L Alkalinity = 200 mg/L as CaCO3 Sulfate ion = 29 mg/L Chloride ion = 14 mg/L pH = 7.7
(a) Calculate the chemical doses needed for excess lime softening. Draw a bar graph for the finished water after two-stage precipitation softening by excess lime treatment with intermediate and final recarbonation. Assume that half the alkalinity in the finished water is
in the bicarbonate form.
(b) Calculate the lime dosage required for selective calcium carbonate removal. Draw a bar graph for the finished water. Is this softening process recommended for this water?
Preview problems
1.... Identify the correct description about Fe & Mn removal, mixing, coagulation, and flocculation
in water treatment.
(a). Colloids are not very stably suspended in water
(b). Adding either alum or Fe salts will increase alkalinity
(c). The purpose of aeration in Fe & Mn removal is to strips out dissolved gases and adds oxygen
(d). Majority of manganese dioxide and iron oxide removed in the sedimentation tank
2...Coagulation of a soft water requires 40 mg/L of alum plus lime to supplement the natural alkalinity for good floc formation. It is desired to react only 10 mg CaCO3/L of the natural alkalinity. What dosage of lime (mg/L as CaO) is required?
3.... A dosage of 30 mg/L of alum and a stoichiometric amount of soda ash are added in coagulation of a surface water. What changes takes place in the ionic character of the water as a result of this chemical treatment?

Explanation / Answer

Q.1:

Cations

Ion

Molar mass

Charge

Conc(mg/L)

Eqv(meq/L)

Ca2+

40

2

175

8.75

Mg2+

24.3

2

40

3.29

Na+

23

1

14

0.61

K+

39.1

1

4

0.10

Total meqv/L of cations = 8.75 + 3.29 + 0.61 + 0.10 = 12.75

Anions:

Ions

Molar mass

charge

Conc(mg/L)

Eqv(meq/L)

Alk(HCO3-)

50

1

200

4.0

Sulfate ion

96

2

29

0.60

Chloride ion

35.5

1

14

0.39

Total meqv/L of anions = 4.0 + 0.60 + 0.39 = 4.99

The milliequivalents-per-liter diagram is shown in the figure below:

Ion

Molar mass

Charge

Conc(mg/L)

Eqv(meq/L)

Ca2+

40

2

175

8.75

Mg2+

24.3

2

40

3.29

Na+

23

1

14

0.61

K+

39.1

1

4

0.10