Download Chapter 4 Aqueous Reactions and Solution Stoichiometry

CHE141
Chapter 4
Chapter 4
Aqueous Reactions and Solution Stoichiometry
1. The total concentration of ions in a 0.250 M solution of HCl is __________.
(a). 1.000 M
(b). 0.125 M
(c). 0.500 M
(d). 0.0625 M
Explanation: Since HCl is a strong acid, it will dissociate completely in solution
according to the equation: HCl
H+ + Cl-. Since the mole:mole ratio between
+
the HCl and the H and Cl ions is 1:1, a 0.250 M solution on dissociation will
produce 0.250 M solution of H+ and 0.250 M solution of Cl-. Thus the sum of thee
two will be equal to 0.500 M
2. A strong electrolyte is one that __________ completely in solution.
(a). reacts
(b). decomposes
(c). disappears
(d).ionizes
3. A weak electrolyte exists predominantly as __________ in solution.
(a). atoms
(b). ions
(c). free radicals
(d).molecules
4. Which of the following are strong electrolytes?
HC1, HC2H3O2, NH3, KC1
(a). HC1, NH3, NaC1
(b).HC1, NaC1
(c). HC1, HC2H3O2, NH3, NaC1
(d). HC1, HC2H3O2, NaC1
Explanation: HCl is a strong electrolyte by definition while since NaCl is a water
soluble ionic salt it will dissolve completely, making it a strong electrolyte also.
Copyright © 2006 Dr. Harshavardhan D. Bapat
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CHE141
Chapter 4
5. Which of the following are weak electrolytes?
HC1, HC2H3O2, NH3
(a). HC1, NH3
(b). HC2H3O2, HC1
(c). HC2H3O2, NH3
(d). All of these
6. Which of the following are strong acids?
HI, HNO3, HF, HBr
(a). HF, HBr
(b).HI, HNO3, HBr
(c). HI, HF, HBr
(d). HNO3, HF, HBr
7. Which hydroxides are strong bases?
Sr(OH)2, KOH, NaOH, Ba(OH)2
(a). KOH, Ba(OH)2
(b). KOH, NaOH
(c). KOH, NaOH, Ba(OH)2
(d) Sr(OH)2, KOH, NaOH and Ba(OH)2
8. A neutralization reaction between an acid and a metal hydroxide produces
__________.
(a). water and a salt
(b). hydrogen gas
(c). precipitate
(d). a sodium salt
9. What is the concentration (M) of KC1 in a solution made by mixing 25.0 mL of 0.100
M KC1 with 50.0 mL of 0.100 M KC1?
(a). 0.0500
(b).0.100
(c). 0.0333
(d). 0.0250
Explanation: Since the 2 solutions have the same composition and concentration to
begin with, it will not change the concentration of the final 75.0 ml of solution.
Copyright © 2006 Dr. Harshavardhan D. Bapat
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CHE141
Chapter 4
10. What is the concentration (M) of CH3OH in a solution prepared by dissolving 11.7 g
of CH3OH in sufficient water to give exactly 230 mL of solution?
(a). 11.7
(b). 2.30 x 10-2
(c). 0.0841
(d).1.59
Explanation: Need to convert the grams of CH3OH to moles and then find the
molarity of the solution by using the molarity formula. Do not forget to convert the
ml to L.
1 mole CH 3 OH
1
11.7 g CH 3 OH !
!
= 1.59 M
32.042 g
0.230 L
11. How many grams of H3PO4 are in 35.1 mL of a 2.75 M solution of H3PO4?
(a). 0.61
(b).9.46
(c). 20
(d). 4.9
Explanation: Need to convert the ml of H3PO4 to liters and then find the # of moles
of phosphoric acid. The moles of phosphoric acid can then be converted to grams of
phosphoric acid.
1L
97.99 g H 3PO 4
35.1 ml !
! 2.75 M !
= 9.46 g H 3PO 4
1000 ml
1 mole
12. What is the concentration (M) of a Na2SO4 solution prepared by dissolving 5.35 g of
Na2SO4 in sufficient water to give 330 mL of solution?
(a). 1.14 x 102
(b). 0.016
(c). 61.7
(d).0.114
Explanation: Convert grams of Na2SO4 to moles of Na2SO4, ml of water to liters of
water and then find the molarity of the solution by using the molarity formula.
5.35 g Na 2SO 4 !
1 mole Na 2SO 4
1
!
= 0.114 M Na 2SO 4
142.035 g Na 2SO 4 0.330 L
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CHE141
Chapter 4
13. How many grams of LiOH are there in 750.0 mL of a 0.0158 M LiOH solution?
(a). 2.11 x 10-5
(b). 11.3
(c). 0.284
(d). 3.50
Explanation: Calculate the number of moles of LiOH present in this solution using
the molarity formula and then convert the number of moles to grams of LiOH.
23.948 g
7.50 ! 10"1 L ! 0.0158 M !
= 0.284 g LiOH
1 mole LiOH
14. How many moles of the iodide ions are present in 45.6 ml of a 4.00 x 10-3 M solution
of Cobalt (II) iodide?
(a). 2.00
(b). 0.500
(c). 0.160
(d).3.65 x 10-4
Explanation: When CoI2 dissolves in water it will ionize completely to produce
cobalt and iodide ions according to the equation: CoI2
Co2+ + 2I-,
producing 2 moles of iodide ions per CoI2 mole. Calculate the number of moles of
CoI2 present in the solution first and then factor in the stoichiometric ratio between
the iodide and the CoI2.
2 moles I !
4.56 " 10! 2 L " 4.00 " 10! 3 M "
= 3.65 " 10! 4 moles iodide
1 mole CoI 2
15. How many moles of K+ are present in 34.3 mL of a 1.27 M solution of K3PO4?
(a). 0.436
(b).0.131
(c). 1.27
(d). 0.0343
Explanation: When K3PO4 dissolves in water it will ionize completely according to
the equation: K3PO4
3K+ + PO43-, producing 3 moles of K+ ions every mole of
K3PO4. Calculate the number of moles of K3PO4 present and then factor in the
stoichiometric ratio between potassium and the K3PO4.
3.43 " 10! 2 L " 1.27 M "
3 moles K +
= 0.131 moles of K +
1 mole K 3PO 4
Copyright © 2006 Dr. Harshavardhan D. Bapat
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CHE141
Chapter 4
16. What are the respective concentrations (M) of Na+ and SO42- produced by dissolving
0.500 mol Na2SO4 in water and diluting to 1.33 L?
(a). 0.665 and 0.665
(b). 0.665 and 1.33
(c). 0.752 and 0.376
(d) 0.376 and 0.752
Explanation: Calculate the molarity of the solution and then the molarities of
individual ions by factoring in the stoichiometric ratios between each ion and the
sodium sulfate. When sodium sulfate dissolves in water each mole of sodium sulfate
will generate 2 moles of sodium and one mole of sulfate ions. Na2SO4
2Na+
2+ SO4
2 moles Na +
1
0.500 moles Na 2SO 4 !
!
= 0.752 moles Na +
1 mole Na 2SO 4 1.33 L
1 mole SO 24 1
0.500 moles Na 2SO 4 !
!
= 0.376 moles SO 24 1 mole Na 2SO 4 1.33 L
17. How many ml of 0.875 M NaOH will be required to completely titrate 25.00 ml of
0.870 M HCl?
(a) 25.00 ml
(b) 24.9 ml
(c) 49.9 ml
(d) None of the above
Explanation: Do not use the M1V1 = M2V2 formula here. Need to write a balanced
equation for the reaction between NaOH and HCl and then calculate the ml of NaOH
needed by using the stoichiometric relationships between the 2 reactants and the
molarity formula.
HCl + NaOH
25.00 ml ??
0.870 M 0.875 M
2.5 ! 10" 2 LHCl ! 0.870MHCl !
NaCl + H2O
1moleNaOH
1L
1000ml
!
!
1moleHCl
0.875MNaOH
1L
= 24.9mlNaOH
18. How many ml of 0.875 M NaOH will be required to completely titrate 25.00 ml of
0.870 M H2SO4?
(a) 25.00 ml
(b) 49.9 ml
(c) 49.7 ml
(d) None of the above
Copyright © 2006 Dr. Harshavardhan D. Bapat
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CHE141
Chapter 4
Explanation: Do not use the M1V1 = M2V2 formula here. Need to write a balanced
equation for the reaction between NaOH and H2SO4 and then calculate the ml of
NaOH needed by using the stoichiometric relationships between the 2 reactants and
the molarity formula. 2NaOH + H2SO4
Na2SO4 + 2H2O
2.50 ! 10" 2 L ! 0.870 MH 2 SO4 !
2molesNaOH
1
1000ml
!
!
= 49.7 ml
1moleH 2 SO4 0.875M
1L
19. The molarity of a solution prepared by diluting 43.72 mL of 5.005 M aqueous
K2Cr2O7 to 500 mL is __________.
(a). 57.2
(b). 0.0044
(c). 0.438
(d). 0.0879
Explanation: You should use the M1V1 = M2V2 formula here. It is the only formula
that you should use here since this is just a dilution of the same substance and no
chemical reactions are involved. Change the volumes to liters on both sides to get the
correct answer.
20. Consider the following 0.100 M solutions. Which one of these solutions will have the
highest concentration of sodium ions?
(a). Sodium phosphate
(b). Sodium sulfate
(c). Sodium hydroxide
(d). Sodium nitrate
Explanation: Knowing the formulas of these compounds allows you to find the
answer here. The formulas are Na3PO4, Na2SO4, NaOH and NaNO3 respectively and
the Na3PO4 has the largest moles of sodium ions per mole of compound compared to
all others.
Copyright © 2006 Dr. Harshavardhan D. Bapat
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