Page 193 - Basic physical chemistry for the atmospheric sciences

Answers, hints, and solutions to exercises 1 7 ')

CHAPTER 4
4. 1 0. 2 . 1 % ; 0.21 M; . 2 4 m ; 0 . 0 1 9
0
4. 1 1 . Hint: Apply LeChatelier's principle.
4. 1 2 . 1 . 5 6 m . Hint: Apply Henry' s law.
4. 13 . 2 . 3 7 kg. Hint: Apply Raoul ' s law.
t
4. 14. Three parts of glycol to 10 parts of water. Hint: Apply
Eq. (4.7).
i
4. 1 5 . Ba(OH)i, KN0 , H 2 S04, and HCI are soluble n water; the
3
remaining have low solubilities . Hint: See Table 4. 1 .
4. 16. 660 kJ.
4. 1 7 . (a) [Ag + (aq)]2[Cro � - (aq) ] . (b) [Ca 2 + ( aq)] [SO l - ( aq)] . (c)
[H + (aq)][C H 3Coo - ( aq)] .
4 . 1 8 . 5 . 3 x I 0 - 3 g L - 1 •
Solution:

CaC0 (s � ) Ca2 + (aq) + Co� - (aq)
3
Therefore, for each mole of CaC03(s) that dissolves in wa­
ter, 1 mole of Ca2 + (aq) and I mole of co� - (aq) enter the
solution. Let x be the solubility of CaC03(s) in moles per
2
liter, then the molar concentrations of Ca + (aq) and
co� - (aq) will each be x. Therefore, since,

K,P = [Ca2 + (aq)][CO� - ( aq)] = 2 . 8 x 1 0 -9

[x] 2 = 2 . 8 x 1 0 - 9
or,
[x] = 5 . 2 9 X 1 0 - 5

Hence, the solubility of CaC03(s) i s 5 . 2 9 x 10 - 5 M . There­
fore, 5 . 29 x 1 0 - 5 moles of CaCOJ{s) dissolves in 1 L of
water. Or, I mole of CaC0 3 (s) dissolves in 1 / (5. 2 9 x 1 0 - 5 )
or 1 . 89 x 1 0 4 L of water. Since the molecular weight of
CaC03 i s 1 0 0. 1 , 1 0 0. I g of CaC03(s) dissolve in 1 . 89 x 1 0 4 L
J O O . I
of water. Therefore , or . 29 1 0 - 3 g of CaC03(s)
5
x
1 . 89 x 1 0 4
dissolves in l L of water, which is therefore the solubility of
CaC03(s) in water.
4. 1 9 . 4 . 3 x 1 0 - 1 3 M .
4 . 2 0. (a) . 3 x 1 0 - 4 M. (b) 1 . 6 x 1 0 - 7 M . Hint: Lead nitrate acts as
l
2
a source of Pb + ( aq) ions in the solution, as does PbS04(s).