CIVE 3610 – Water Supply and Treatment
Fall Semester 2004
Homework #2 - Solution
1. What are the molecular weight and equivalent weight of CaCl2? If a water sample
contains 168mg/L of CaCl2, how many milliequivalents are present?
Solution:
CaCl2 → Ca2+ +2ClValence = 2
Molecular weight = 40 + 2*(35.5) = 111 g/mole
Equivalent Weight = (111g/mole) / (2 eq/mole) = 55 g/eq
Milliequivalents of Cacl2 present in 168 mg/l of CaCl2 = (168mg/L)* (1meq/55mg)
= 3.03 meq/L
2. How much water must be added to 100mL of a 0.75 molar solution of KCL to
make a 0.04 molar solution?
Solution:
C1V1 = C2V2
(0.75M) * (100mL) = (0.04M) * (X mL)
i.e., X =1875 mL
Quantity of water to be added = (1875mL -100mL) = 1775mL
3. What mass of lead nitrate must be dissolved in 1L of water to produce a solution
that contains 20 mg of lead ions? Assume 100% ionization.
Solution:
Lead, Pb = 207g/mole
Lead Nitrate, PbNO3 = 207+14+3(16) =269 g/mole
(Pb/ PbNO3) =
207 g Pb/mole
g Pb
= 0.77
269 g PbNO3 /mole
g PbNO3
Therefore, 0.77 =
20 mg Pb
X mg PbNO3
X = 25.9 mg PbNO3
4. Given the following raw water quality data:
Determine:
a. Does electroneutrality hold?
b. Total Hardness (TH), carbonate hardness (CH), noncarbonate hardness
(NCH) in mg/L as CaCO3 and meq/L.
c. Is this water soft, moderately hard, hard, or very hard?
d. Alkalinity in mg/L as CaCO3 and meq/L.
e. H+ and OH- in mg/L.
f. Calculate TDS.
Parameter
Ca2+
Mg2+
Na+
K+
Fe2+
HCO3SO42ClFSiO2
pH
Source Data (mg/L)
9.2
2.6
202
5
0.02
410
2
98
1.4
16
7.9
Solution:
mg/L
Ca2+
Mg2+
Na+
K+
Fe2+
HCO3SO42ClFSiO2
pH
9.2
2.6
202
5
0.02
410
2
98
1.4
16
7.9
MW
EW
meq/L
40.1
24.3
23.0
39.1
55.8
61.0
96.1
35.5
19
20.05
12.15
23.00
39.10
27.90
61.00
48.05
35.5
19.00
0.46
0.21
8.78
0.13
0.00
6.72
0.04
2.76
0.07
mg/L
CaCO3
23.0
10.5
439
6.5
0.0
336.0
2.0
138.0
3.5
Sample Calculation:
CA2+
MW = 40.1 g/mole
EW = MW/2 = 40.1/2 = 20.02 g/eq
meq/L = mg/L/ EW = (9.2/20.05) = 0.46 meq/L
mg/L as CaCO3 = meq/L * EW CaCO3
= 0.46*50 = 230 mg/L as CaCO3
a) Does electroneutrality hold?
(Check with meq/L)
∑ cations = 0.46+0.21+8.78+0.13+0 = 9.58
∑anions = 336 +2.0+138+3.5 =9.59
Since ∑ cations ≈ ∑anions
Electroneutrality holds.
b) Total hardness (TH), carbonate hardness (CH), noncarbonate hardness (NCH) in
mg/L as CaCO3 and meq/L.
TH = ∑ Mg2+ + Ca2+
TH in mg/l as CaCO3 = 23.0+10.5 = 33.5 mg/L CaCO3
TH in meq/L = 0.46+0.21 = 0.67 meq/L
CH = ∑ HCO3CH in mg/l as CaCO3 = 336 mg/L CaCO3
CH in meq/L = 6.72 meq/L
NCH = TH-CH
As CH>TH, TH = CH
i.e., TH = 336 mg/L as CaCO3 or 6.72 meq/L
Therefore, NCH = 0
c) Is this water soft, moderately hard, hard, or very hard?
Water is very hard. (As CH = 336 mg/L CaCO3 > 300 mg/L CaCO3)
d) Alkalinity in mg/L as CaCO3 and meq/L.
Alkalinity = CH = 336 mg/L CaCO3 or 6.72 meq/L
e) H+ and OH- in mg/L.
[H+] = (10-7.9 moles/L) * (1g/mole H+)*(1000 mg/1g) = 1.3*10-5 mg/L H+
[OH-] = (10-6.1 moles/L) * (17g/mole OH-)*(1000 mg/1g) = 1.3*10-2 mg/L OHf) Calculate TDS
TDS = sum of all ions in mg/L
= 9.2+2.6+202+5+0.02+410+2+98+1.4+16+7.9
= 730.22 mg/L