Chemistry for Environmental Engineering and Science, 5th Edition

1



Chemistry for Environmental Engineering and Science, 5th Edition

CHAPTER 2 - PROBLEM SOLUTIONS

2.1 (a) MgCO3

FW = 24.3 + 12 + 3(16) = 84.3 g/mol
MgCO3 + 2 H+ = Mg2+ + H2CO3
84.3
EW = 2 = 42.15 g/equiv
(b) NaNO3
FW = 23 + 14 + 3(16) = 85 g/mol
NaNO3 + H+ = Na+ + HNO3
85
EW = 1 = 85 g/equiv

(c) CO2
FW = 12 + 2(16) = 44 g/mol
CO2 + H2O = H2CO3
2–
Note from above: MgCO3 + 2H+ = Mg2+ + H2CO3 and H2CO3 = 2H+ + CO 3
or: CaCO3 + 2H+ = Ca2+ + H2CO3
44
EW = 2 = 22 g/equiv
* + –
Note: In some reactions, Z might be considered to be 1 (i.e., H2CO3 = H + HCO3 )
(d) K2HPO4
FW = 2(39.1) + 1 + 31 + 4(16) = 174.2 g/mol
K2HPO4 + 2H+ = H3PO4 + 2K+
174.2
EW = = 87.1 g/equiv
2


2.2 (a) BaSO4

FW = 137.3 + 32.1 + 4(16) = 233.4 g/mol
BaSO4 + 2 H+ = Ba2+ + H2SO4
233.4
EW = = 116.7 g/equiv
2
(b) NaCO3



2-1

, 2



FW = 2(23) + 12 + 3(16) = 106 g/mol
NaCO3 + 2 H+ = Na+ + H2CO3
106
EW =
2 = 53 g/equiv


(c) H2SO4

FW = 2(1) + 32.1 + 4(16) = 98.1 g/mol
2–
H2SO4 = 2H+ + SO4
98.1
EW =
2 = 49.05 g/equiv
(d) Mg(OH)2

FW = 24.3 + 2(16) + 2(1) = 58.3 g/mol
Mg(OH)2 + 2 H+ = Mg2+ + 2H2O
58.3
EW = 2 = 29.15 g/equiv


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2.3 (a) 23+17 = 0.25 for NaOH
10 10
(b) 46+32+64 = 142 = 0.0704 for Na2SO4
10 10
(c) 78+104+7(16) = 294 = 0.034 for K2Cr2O7
10 10
(d) 39+35.5 = 74.5 = 0.134 for KCl.

X
2.4 (a) 2 = 0.15 M, X = 0.30 mol KMnO4

FW = 39.1 + 24.3 + 4(16) = 127.4 g/mol
0.30(127.4) = 38.22 g
X
(b) 2 = 0.15 N, X = 0.30 equiv. KMnO4

127.4
EW = = 63.7 g/equiv
2

0.30(63.7) = 19.11 g




2-2

, 3


40
2.5 Ca2+: EW = 2 = 20 g/equiv
44
meq/l = 20 = 2.2 meq/L

24.3
Mg2+: EW = 2 = 12.15 g/equiv
19
meq/l = 12.15 = 1.56 meq/L

Total Hardness = 2.20 + 1.56 = 3.76 meq/L
= 3.76(50 mg/meq) = 188 mg/lL as CaCO3


2.6 – + –
Note: for HCO3 , H , and OH , mol/L = equiv/L (Z = 1)
2–
for CO3 , equiv/L = 2(mol/L) (Z = 2)
[OH–][H+] = 10-14 F|} for [H+] = 10-9.5, [OH–] = 10-4.5
pH = –log [H+] |J
– 118 mg/L -3
[HCO3 ] = 61,000 mg/mol = 1.93 x 10 M
2– 19 mg/L -4
[CO3 ] = 60,000 mg/mol = 3.17 x 10 M

equiv/L Alk = 1.93 x 10-3 + 2(3.17 x 10-4) + 10-4.5 – 10-7.5

Alk = 2.60 x 10-3 equiv/L (50,000 mg/equiv)
Alk = 130 mg/L as CaCO3


2.7 (a) CaCl2 + Na2CO3  CaCO3(s) + 2 NaCl
(b) Ca3(PO4)2 + 4 H3PO4  3 Ca(H2PO4)2
(c) MnO2 + 2NaCl + 2H2SO4  MnSO4 + 2H2O + Cl2 + Na2SO4
(d) Ca(H2PO4)2 + 2NaHCO3  CaHPO4 + Na2HPO4 + 2H2O + 2CO2


2.8 (a) FeS + 2HCl  FeCl2 + H2S
(b) 3Ca2 + 6KOH  5KCl + KClO3 + 3H2O
(c) 6FeSO4 + K2Cr2O7 + 7H2SO4  3Fe2(SO4)3 + Cr2(SO4)3 + K2SO4 + 7H2O
(d) Al2(SO4)3 • 14H2O + 3Ca(HCO3)2  2Al(OH)3 + 3CaSO4 + 14H2O + 6CO2


2.9 (a) 4Fe(OH)2 + 2H2O + O2  4Fe(OH)3



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