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2.1 2.14 The solutions for these problems are the solutions for
problems 2.1 - 2.14 in
the 2nd
edition Solutions Manual.
2.15EDTA (C10N2O8H16) has a molecular weight of 292 g/mole
Calcium molar concentration: (20 mg/L)(40.1 mol/g)-1
(103
mg/g)-1
= 0.000499 M
Mass of EDTA: (0.000499 mol/L)(292 g/mol)(44 gal)(3.785 L/gal) =
24.3 g EDTA
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2.16 The balanced equation is: 2Al0 + 6HCl 2AlCl3 + 3H2
(g)So,
2
0
2
2
0
0
2
0
Hg
Alg9
Hg2
Hmole
Almole
Alg27
Hmoles3
Almoles2=
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2.17 - 2.21 The solutions for these problems are the solutions
for problems 2.15 - 2.19 in
the 2nd
edition Solutions Manual.
2.22 From Table 2.3, Ksp,CaSO4 is 2 x 10-5 and Ksp = [Ca2+]
[SO42-] = [SO42-]2At saturation: [SO4
2-] = Ksp
= (2 x 10
-5)
= 0.00447 mol/L
Check if saturation is exceeded: mol/L0.00368g136
mole
L
g5.0=
Since saturation is not exceeded the sulfate concentration is
0.00368 mol/L
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2.23 Assume a closed system.
a. [Ca2+
] = Ctot = [H2CO3] + [HCO3-] + [CO3
2-] eqn. 1
Charge balance: [H+] + 2[Ca
2+] = [HCO3
-] + 2[CO3
2-] + [OH
-] eqn. 2
At pH 8.5: Ctot [HCO3-] [CO3
2-] (see Figure 2.5) and [H
+] [OH
-]
So eqn. 2 simplifies to: 2[Ca2+
] = Ctot + [OH-] eqn. 3
Using eqn. 1 in eqn. 3 yields: [Ca2+
] = [OH-] = 10
-5.5= 3.16 x 10
-6
Note this is less than the Ca
2+initially added, so CaCO3(s) has precipitated
b.
[ ] [ ] ( )( )
( )M10M102.24
104.47
1010
HCOHCO 7.65--8
7-
-5.5-8.5
1
3(aq)2 ==
==+
K
c.
[ ] ( )( ) g/L0.20g/L0.19968mol
g100M10-M102CaCO 5.5-3-(s)3 ==
=
2.24 - 2.36 The solutions for these problems are the solutions
for problems 2.20 - 2.32 in
the 2
nd
edition Solutions Manual.
