Chemistry for Environmental Engineering and Science, 5th Edition

Chemistry for Environmental Engineering and Science, 5th Edition


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

,Chemistry for Environmental Engineering and Science, 5th Edition


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
H2SO4 = 2H+ + SO2–
4
98.1
EW = = 49.05 g/equiv
2
(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


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

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

2-2

, Chemistry for Environmental Engineering and Science, 5th Edition

0.30(63.7) = 19.11 g




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