Exercício Resolvido - Quimica Geral (86)

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Chapter 5: Introduction to Reactions in Aqueous Solutions 
 171 
 (c) 
3+
2 43+ 3+3
2 4 3
0.083 mol Al SO 2 mol Al
Al = = 0.166 M Al
1 L soln 1 mol Al SO
 
(d) Na
 mol Na PO
 L soln
 mol Na
 mol Na PO
 M Na+ 3 4
+
3 4
+=
0.209
1
3
1
= 0.627 
 
8. (E) Choice (d) is the solution with the greatest concentration of sulfate ions. 
 (a) 
2-
2-4
2 4 4
2 4
1 mol SO
0.075 M H SO 0.075 M SO
1 mol H SO
 
 (b) 
2-
2-4
4 4
4
1 mol SO
0.22 M MgSO 0.22 M SO
1 mol MgSO
 
 (c) 
2-
2-4
2 4 4
2 4
1 mol SO
0.15 M Na SO 0.15 M SO
1 mol Na SO
 
 (d) 
2-
2-4
2 4 3 4
2 4 3
3 mol SO
0.080 M Al (SO ) 0.24 M SO
1 mol Al (SO )
 
 (e) 
2-
2-4
4 4
4
1 mol SO
0.20 M CuSO 0.20 M SO
1 mol CuSO
 
 
9. (E) 
Conversion pathway approach: 
2 22 2
2 22 2
0.132 g Ba OH 8H O 1 mol Ba OH 8H O1000 mL 2 mol OH
OH =
275 mL soln 1 L 315.5g Ba OH 8H O 1 mol Ba OH 8 H O
 = 3.04 10 3 M OH 
 
 Stepwise approach: 
 
22
22
22
22
22
2 -3 -2
0.132 g Ba OH 8H O 1000 mL
275 mL soln 1 L
1 mol Ba OH 8H O
315.5g Ba OH 8H O
2 mol OH
1 mol Ba OH 8 H O
 = 0.480 g/L
0.00152 mol Ba OH ×8H O0.480 g
 = 
L L
0.00152 mol Ba OH ×8H O
= 3.04 10 M OH
L
 
 
10. (E) 
 K
 mol KCl
 L soln
 mol K
 mol KCl
 M+
+
=
0.126
1
1
1
= 0.126 K+ 
Mg
 mol MgCl
 L soln
 mol Mg
 mol MgCl
 M2+ 2
2+
2
=
0.148
1
1
1
= 0.148 Mg2+ 
Now determine the amount of Cl in 1.00 L of the solution. 
Chapter 5: Introduction to Reactions in Aqueous Solutions 
172 
2
2
0.148 mol MgCl0.126 mol KCl 1 mol Cl 2 mol Cl
mol Cl = +
1 L soln 1 mol KCl 1 L soln 1 mol MgCl
= 0.126 mol Cl + 0.296 mol Cl = 0.422 mol Cl
 
Cl
 mol Cl
 L soln
 M Cl=
0.422
1
= 0.422 
 
 
11. (E) 
 (a) 
2+ 2+ 2+
2+ 4 2+
2+ 2+
 14.2 mg Ca 1 g Ca 1 mol Ca
[Ca ] = 3.54 10 M Ca
1 L solution 1000 mg Ca 40.078 g Ca
 
 (b) 
+ + +
+ 3 +
+ +
 32.8 mg K 1 g K 1000 mL solution 1 mol K
[K ] = 8.39 10 M K
100 mL solution 1 L solution 1000 mg K 39.0983 g K 
 
 (c) 
2+ 2+ 2+
2+ 3 2+
6 2+ 2+
 225 g Zn 1 g Zn 1000 mL solution 1 mol Zn
[Zn ] = 3.44 10 M Zn
1 mL solution 1 L solution 1 10 g Zn 65.39 g Zn
 
 
12. (E) 
5 50.9 mg F 1 g 1 mol F 1 mol NaF[NaF] = = 4.7 10 M = 5 10 M NaF
1 L 1000 mg 19.00 g F 1 mol F
 
 
13 (E) In order to determine the solution with the largest concentration of K+, we begin by 
converting each concentration to a common concentration unit, namely, molarity of K+. 
 
+
+2 4
2 4
0.0850 M K SO 2 mol K
0.17 M K
1 L solution 1 mol K SO
 
 
+
+1000 mL solution1.25 g KBr 1 mol KBr 1 mol K 0.105 M K
100 mL solution 1 L solution 119.0023 g KBr 1 mol KBr
 
 
+ + +
+
+ +
1000 mL solution8.1 mg K 1 g K 1 mol K
0.207 M K
1 mL solution 1 L solution 1000 mg K 39.0983 g K
 
 Clearly, the solution containing 8.1 mg K
+ per mL gives the largest K+ of the three solutions. 
 
14. (E) 
(c) NH3 is a weak base and would have an exceedingly low H
+ ; the answer is not 
1.00 M NH3 . HC H O2 3 2 is a very weak acid; 0.011 M HC H O2 3 2 would have a low H
+ . 
H SO2 4 has two ionizable protons per mole while HCl has but one. Thus, H SO2 4 would 
have the highest H+ in a 0.010 M aqueous solution.