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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¯Ââ€ž Fe3+ + 3ClFePO4(s) Ã¯Ââ€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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