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It’s an environmental engineering class. The quiz is focused on mineral solubility processes. There are 5 numerical questions that I need to enter my final answer with two decimal places.

Practice problems (Mineral solubility)

1. (DM 2-28) The solubility product (Ksp) of calcium fluoride (CaF2) is 3Ãƒâ€”10-11 at 25Ã¯â€šÂ°C. Could a

fluoride concentration of 1.0 mg/L be obtained in water that contains 200 mg/L of calcium?

Answer: yes, because the saturation concentration of fluoride is 1.47 mg/L when the calcium

concentration is 200 mg/L

Solution:

Ã¯â€šÂ·

Calculate the molarity of 200 mg/L of Ca2+

200 Ã¢Ë†â€” 10

Ã¯â€šÂ·

5.0 Ã¢Ë†â€” 10

40

Use the [Ca2+] to determine the saturation concentration of FÃ¢Å¸Â¶

2

3 Ã¢Ë†â€” 10

3 Ã¢Ë†â€” 10

3 Ã¢Ë†â€” 10

6.0 Ã¢Ë†â€” 10

5 Ã¢Ë†â€” 10

/

6.0 Ã¢Ë†â€” 10

7.75 Ã¢Ë†â€” 10

7.75 Ã¢Ë†â€” 10

Ã¯â€šÂ·

19

10

1.47

The saturation concentration is 1.47 mg/L. Smallest concentrations are allowable

(unsaturated solution). Therefore, 1.0 mg/L is possible.

Check:

Ã¢â€¹â€º

?

5.26 Ã¢Ë†â€” 10

1 Ã¢Ë†â€” 10

5.0 Ã¢Ë†â€” 10

3 Ã¢Ë†â€” 10

1.39 Ã¢Ë†â€” 10

5.26 Ã¢Ë†â€” 10

so unsaturated

1.39 Ã¢Ë†â€” 10

2. (ML 3.7) Calculate the solubility (in mg/L) of the following ionic compounds. Assume T=25Ã¯â€šÂ°

C. (Hint: Ã¢â‚¬Å“SolubilityÃ¢â‚¬Â in this problem means the special case of dissolving the salt in a liter of

water.) Use the following Ksp values: 9Ãƒâ€”10-12 for Mg(OH)2 and 6Ãƒâ€”10-18 for FeS

(a) Mg(OH)2 Answer: 7.64 mg/L

(b) FeS Answer: 2.15Ãƒâ€”10-4 mg/L

Solution:

Ã¢Å¸Â¶

(a)

2

9 Ã¢Ë†â€” 10

(ML, pg. 49)

that dissolves and there are no other sources of

If x = # mol

2

2

4

9 Ã¢Ë†â€” 10

Page 1 of 6

and

1.31 Ã¢Ë†â€” 10

24.3

2 17

10

7.64

Ã¢Å¸Â¶

(b)

6 Ã¢Ë†â€” 10

x = # mol FeS that dissolve

6 Ã¢Ë†â€” 10

6 Ã¢Ë†â€” 10

(ML, pg. 49)

2.45 Ã¢Ë†â€” 10

2.45 Ã¢Ë†â€” 10

55.85

32.1

10

2.16 Ã¢Ë†â€” 10

3. (DM 2-26 augmented by JJ) Groundwater in Pherric, New Mexico, is very acidic and initially

contains 1.800 mg/L of iron as Fe3+. Use Ksp=2.67Ãƒâ€”10-39 for Fe(OH)3

(a) To what pH must the water be raised in order to precipitate all but 0.30 mg/L of the iron?

The temperature of the water is 25Ã‚Â°C. (Hint: Ã¢â‚¬Å“All but 0.30 mg/LÃ¢â‚¬Â means that there 0.30 mg/L left

in solution after the Fe(OH)3 precipitates.) Answer: pH = 3.35

(b) If the pH was higher than the answer in part (a), state whether the Fe3+ concentration would

be higher or lower than 0.30 mg/L and provide justification.

Solution:

Ã¯â€šÂ·

The precipitation changes the concentration from 1.8 mg/L to 0.30 mg/L where it stops.

At this concentration, the solution is saturated (in equilibrium).

Ã¢â€¡Å’

3

At equilibrium,

2.67 Ã¢Ë†â€” 10

[

. Ã¢Ë†â€”

.

(Value from an older text. Disagrees with MZ table 3.8)

5.372 Ã¢Ë†â€” 10

Ã¯â€šÂ·

To get the pH, first solve for

/

/

2.67 Ã¢Ë†â€” 10

7.921 Ã¢Ë†â€” 10

5.372 Ã¢Ë†â€” 10

10

1.262 Ã¢Ë†â€” 10

7.921 Ã¢Ë†â€” 10

log 1.262 Ã¢Ë†â€” 10

2.9

Ã¯â€šÂ·

If the pH was higher:

Ã¢â€ â€˜,

If the

was higher and

Ã¢â€ â€œ,

Ã¢â€ â€˜

was constant

Page 2 of 6

[

would be lower than the 0.3 mg/L given in this problem.

4. Consider the salt, aluminum hydroxide, which has the following solubility product:

Al(OH)3(s) = Al3+ + 3OH- Ksp = 5Ãƒâ€”10-33

(a) Is a solution containing [Al3+] = 10-13 M and [OH-] = 10-7 M saturated? Answer: No.

(b) At what pH would the solution with the aluminum concentration in part (a) be saturated?

Answer: 7.57

(c) Suppose we raise the pH to 8 and then drop some solid Al(OH)3 into the solution. Would the

Al(OH)3 we added dissolve? Why or why not? Answer: No.

Solution:

(a) Saturated?

Ã¢â€¹â€º

10

10

?

1.0 Ã¢Ë†â€” 10

5.0 Ã¢Ë†â€” 10

Unsaturated

(b) At what pH is it saturated?

5 Ã¢Ë†â€” 10

10

/

10

3.68 Ã¢Ë†â€” 10

log 2.71 Ã¢Ë†â€” 10

3.68 Ã¢Ë†â€” 10

2.71 Ã¢Ë†â€” 10

7.57

(c) Raise pH

If

7.57 then

3.68 Ã¢Ë†â€” 10

(from above)

The system would be supersaturated and the solid

added would not dissolve.

5. (MZ 3.18) What pH is required to reduce a high concentration of a dissolved Mg2+ to 25

mg/L? The solubility product for the following reaction is 10-11.16. Answer: 9.91

Mg(OH)2 (s) = Mg2+ + 2OHSolution:

Ã¢Å¸Â¶

Ã¯â€šÂ·

Solubility product equation at equilibrium

Page 3 of 6

2

Ã¯â€šÂ·

Ã¯â€šÂ·

Express the Mg concentration in mol/L

/

25 Ã¢Ë†â€” 10

1.03 Ã¢Ë†â€” 10

24.3 /

Solve for

Ã¯â€šÂ·

10 .

1.03 Ã¢Ë†â€” 10

Solve for pH

log

/

8.20 Ã¢Ë†â€” 10

10

1.22 Ã¢Ë†â€” 10

8.20 Ã¢Ë†â€” 10

log 1.22 Ã¢Ë†â€” 10

9.91

6. (MZ 3.20) At a wastewater-treatment plant FeCl3(s) is added to remove excess phosphate from

the effluent. Assume that the reactions that occur are

FeCl3(s) Ã¯Ââ€ž Fe3+ + 3ClFePO4(s) Ã¯Ââ€ž Fe3+ + PO43The equilibrium constant for the second reaction is 10-26.4. What concentration of Fe3+ would be

needed to maintain the phosphate concentration below the limit of 1 mg/L of P? (Hint: Assume

all the P is in the PO43- form so weÃ¢â‚¬â„¢re asking for 1 mg/L PO43- as P.) Answer: 6.88Ãƒâ€”10-18 mg/L

Solution:

The equilibrium constant for the second reaction is 10-26.4. What concentration of Fe3+ would be

needed to maintain the phosphate concentration below the limit of 1 mg/L of P? (Hint: Assume

all the P is in the

form so weÃ¢â‚¬â„¢re really asking for 1 mg/L

as P)

Ã¯â€šÂ·

At equilibrium

10 .

concentration is too low the

If the

Ã¯â€šÂ·

Calculate desired [

]

1 mg/L as P means the P part of

1

1

Ã¯â€šÂ·

94.97

3.07

30.97

3.07 Ã¢Ë†â€” 10

94.97 /

Calculate

from

concentration will be greater than desired.

is 1 mg/L

30.97 64

30.97

3.23 Ã¢Ë†â€” 10

equation

10 .

3.23 Ã¢Ë†â€” 10

1.23 Ã¢Ë†â€” 10

Page 4 of 6

1.23 Ã¢Ë†â€” 10

6.88 Ã¢Ë†â€” 10

7. This problem combines alkalinity and mineral solubility in a realistic scenario.

You perform an alkalinity test on a water sample with an initial pH of 9.75. The alkalinity is

12.1 mg/L as CaCO3.

(a) Calculate the CT value. Answer: 1.54Ãƒâ€”10-4 M

(b) Determine whether there is any potential to precipitate CaCO3 in the pipes given a calcium

concentration of 15 mg/L. Justify your answer. Answer: Calculations show that it exceeds the

Ksp, so yes there is a potential.

(c) For the same calcium concentration and CT, determine the potential to precipitate CaCO3 if

the pH were to be lowered to 7. Answer: No, but why? Show calculations to support your

answer.

Solution:

1

(a)

2

1

1

2

10

.

.

10

.

10

.

.

0.791

.

.

.

.

12.1

0.208

.

.

.

2.42 Ã¢Ë†â€” 10

,

2

2

2.42 Ã¢Ë†â€” 10

10

.

0.791 2 0.208

(b) Might CaCO3 precipitate?

It will if

10

.

1.54 Ã¢Ë†â€” 10

15 Ã¢Ë†â€” 10

3.75 Ã¢Ë†â€” 10

40 /

0.208 1.54 Ã¢Ë†â€” 10

3.20 Ã¢Ë†â€” 10

3.75 Ã¢Ë†â€” 10 3.20 Ã¢Ë†â€” 10

1.20 Ã¢Ë†â€” 10

Page 5 of 6

1.20 Ã¢Ë†â€” 10

5.0 Ã¢Ë†â€” 10

So there is potential for precipitation.

(c) What if we lowered the pH to 7

This changes

and

.

1

1

1

1

3.82 Ã¢Ë†â€” 10

10

10 . 10

1.54 Ã¢Ë†â€” 10

.

10

10

5.88 Ã¢Ë†â€” 10

3.82 Ã¢Ë†â€” 10

.

(much smaller)

Now

3.75 Ã¢Ë†â€” 10

2.21 Ã¢Ë†â€” 10

Because

5.88 Ã¢Ë†â€” 10

2.21 Ã¢Ë†â€” 10

5.0 Ã¢Ë†â€” 10

, the solution is unsaturated and there wonÃ¢â‚¬â„¢t be any precipitation.

Page 6 of 6

Atomic Weights and Ion Charges of Selected Elements

H+ = 1

O= 16 Cl- = 35.5 A13+ = 27

C= 12

P= 31

Nat = 23 K+ = 39.1

N= 14

S = 32

Mg2+ = 24.3

Ca2+ = 40

Fe2+ or Fe 3+ = 55.85

Cu2+ = 63.5

= 65.4

F+ = 19

Pb2+ = 207.2

Hg2+ = 200.6

Zn2+

Conversion Factors and Constants

7.48 gal/ft?

3.785 L/gal

y = 62.4 lb/ft?, H20 3.28 ft/m

N=kg m/s2 W = Nm/s

2.205 lb/kg

43,560 ft/ac

454 g/lb

104 m2/ha

50,000 mg CaCO3 per eq.

1000 L/m3

273Ã‚Â°K = 0Ã‚Â°C

Pa = N/m2

1 mg/L = 1 g/m

14.7 psia/atm

101 kPa/atm

Henry’s Law Constants, 104 atm

TÃ‚Â°C CO2 N2 02

0 0.0728 5.29 2.55

10 0.104 6.68 3.27

20 0.142 8.04 4.01

CH4

2.24

2.97

3.76

Henry’s Law Constants at 25Ã‚Â°C, atm L/mol

Chloroform

4.1

Tetrachloroethene

26.9

Benzene

5.6

Hydrogen sulfide (H2S)

10

10.33

Acid Dissociation Constants, pK, values

Carbonic acid (H2CO3) 6.35

Acetic acid (CH3COOH) 4.75

Hydrogen sulfide (H2S)

7.2

Hypochlorous acid (HOCI)

7.54

Oxalic acid (H2C204)

1.27

11.89

Hydroflouric acid (HF)

Nitric acid (HNO3)

Ammonium (NH4+)

Hydrochloric acid (HCI)

Phosphoric acid (H3PO4)

3.4

0

9.3

-3

2.15

4.27

7.20

12.4

Solubility Products, Ksp at 25Ã‚Â°C

CaCO3

4.96×10-9

Ca(OH)2

4.68×10-6

Cu(OH)2

2.2×10-20

Fe(OH)3

2.79×10-39

Al(OH)3

5.0*10-33

Fes

CaF

Pb(OH)2

Mg(OH)2

MgCO3

1.57×10-19

3.0×10-11

1.40×10-20

1.82×10-11

1.15×10-5

Properties of Water (use linear interpolation for temperatures not shown)

Temperature (Ã‚Â°C), T

5

10

15

Viscosity (Pa-s), u

1.519×10-3 1.307×10-3 1.139×10-3

Specific weight (N/m?), y

9807

9804

9798

Density (kg/mÃ‚Âº), p

1000

999.7

999.1

20

1.002×10-3

9789

998.2

Temperature adjustment coefficient (0) for k:

0 = 1.135 for temp = 4 Ã¢â‚¬â€œ 20Ã‚Â°C

0 = 1.056 for temp = 20 Ã¢â‚¬â€œ 30Ã‚Â°C

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