Chem 020 Tutorial Test 2, Section 004
Monday October 15, 07/08
Name:
TA:
1. What volume, in liters, of 0.3 M KOH is required to neutralize 0.5 liters of 0.7 M
H2SO4?
H2SO4 is diprotic, therefore Moles KOH required = 2 x moles H2SO4
Moles H2SO4 = MV = 0.7 x 0.5 = 0.35
Moles KOH required = 2 x 0.35 = MKOH x VKOH = 0.3 x VKOH
VKOH = 2.33 liters
1 mark for recognizing H2SO4 is diprotic, 1 mark for the correct answer.
2. A bottle of concentrated HCl has a density of 1.70 g/ml. When 1.0 ml of this
concentrated solution is diluted to 1.5 liters the pH of the resulting solution is found to be
2.0. Find: a) the Molarity of the HCl in the original concentrated solution and b) the % by
mass of HCl in the original concentrated solution.
If pH = 2 then [HCl] = .01
The moles HCl in the diluted solution may be obtained from:
Moles HCl = Mdilute solution x Vdilute solution = 0.01 x 1.5 = 0.015
Therefore we know 1.0 ml of the original concentrated solution contained 0.015 moles
HCl and:
MHCl = moles/volume = 0.015/.001 = 15
Now we also know 1 ml of the original concentrated solution contains 0.015 moles HCl (
mass = moles x molar mass = 0.015 x 36.45 = 0.547 g and, from the density, we know I
ml of this solution weighs 1.70 g.
% HCl by mass = [mass HCl/mass solution] x 100 = 32.2 %
1mark for concentration HCl in the dilute solution, 1 additional mark for the rest.
3. A sample of an unknown triprotic acid weighing 9.8 grams is neutralized by 750 ml 0f
0.4 M NaOH. Find the molar mass of the acid.
3 moles of NaOH are required for each mole of the acid.
Moles NaOH = MV = 0.4 x .750 = 0.3
Therefore moles acid = 0.3/3 = 0.1
Mass = moles x Molar mass 9.8 = 0.1 x Molar mass
Molar mass = 98g
2 marks for correct answer.
4. Balance the following in acidic solution:
Cu + NO3- → Cu2+ + NO2
Separate into ox and red and balance for elements except H and O
Cu → Cu2+ (Oxidation)
NO3- → NO2 (Reduction)
Assign oxidation #s and balance for electrons:
Cu → Cu2+ + 2eNO3- + e- → NO2
+
Balance charge with H
Cu → Cu2+ + 2e+
2H + NO3- + e- → NO2
Balance O with H2O
Cu → Cu2+ + 2e2H+ + NO3- + e- → NO2 + H2O
Multiply be factor to eliminate 2e- and add
Cu → Cu2+ + 2e4H + 2 NO3 + 2 e- → 2 NO2 + 2 H2O
+
-
Cu + 4H+ + 2 NO3- → Cu2+ + 2 NO2 + 2 H2O
2 marks for correct answer (this example was done in class)
1.
Balance the following in basic solution:
Zn + NO3- → Zn(OH)42- + NH3
Separate into ox and red and balance for elements except H and O
Zn → Zn(OH)42- + (Oxidation)
NO3- → NH3 (Reduction)
Assign oxidation #s and balance for electrons:
Zn → Zn(OH)42- + 2eNO3- + 8 e- → NH3
Balance charge with OH
4OH- + Zn → Zn(OH)42- + 2eNO3- + 8 e- → NH3 + 9 OH-
Balance O with H2O
4OH- + Zn → Zn(OH)42- + 2e6 H2O + NO3- + 8 e- → NH3 + 9 OH-
Multiply be factor to eliminate e- and add
16OH- + 4Zn → 4 Zn(OH)42- + 8e6 H2O + NO3- + 8 e- → NH3 + 9 OH7OH- + 4Zn + 6 H2O + NO3-
→ 4 Zn(OH)42- + NH3 +
One mark for correct separation into ox and red. One additional mark for correct
answer.
Chem 020 Tutorial Test 2, Section 001
Tuesday October 16, 07/08
Name:
TA:
1. What volume, in liters, of 0.3 M KOH is required to neutralize 0.5 liters of 0.7 M
H2SO4?
H2SO4 is diprotic, therefore Moles KOH required = 2 x moles H2SO4
Moles H2SO4 = MV = 0.7 x 0.5 = 0.35
Moles KOH required = 2 x 0.35 = MKOH x VKOH = 0.3 x VKOH
VKOH = 2.33
1 mark for recognizing H2SO4 is diprotic, 1 mark for the correct answer.
2. A bottle of concentrated HCl has a density of 1.70 g/ml. When 1.0 ml of this
concentrated solution is diluted to 1.5 liters the pH of the resulting solution is found to be
2.0. Find: a) the Molarity of the HCl in the original concentrated solution and b) the % by
mass of HCl in the original concentrated solution.
If pH = 2 then [HCl] = .01
The moles HCl in the diluted solution may be obtained from:
Moles HCl = Mdilute solution x Vdilute solution = 0.01 x 1.5 = 0.015
Therefore we know 1.0 ml of the original concentrated solution contained 0.015 moles
HCl and:
MHCl = moles/volume = 0.015/.001 = 15
Now we also know 1 ml of the original concentrated solution contains 0.015 moles HCl (
mass = moles x molar mass = 0.015 x 36.45 = 0.547 g and, from the density, we know I
ml of this solution weighs 1.70 g.
% HCl by mass = [mass HCl/mass solution] x 100 = 32.2 %
1mark for concentration HCl in the dilute solution, 1 additional mark for the rest.
3. A sample of an unknown diprotic acid weighing 13.8 grams is neutralized by 750 ml 0f
0.4 M NaOH. Find the molar mass of the acid.
2 moles of NaOH are required for each mole of the acid.
Moles NaOH = MV = 0.4 x .750 = 0.3
Therefore moles acid = 0.3/2 = 0.15
And Mass = moles x Molar mass
13.8 = 0.15 x MM MM= 92 g
2 marks for correct answer.
4. Balance the following in acidic solution:
Cu + SO42- → Cu2+ + SO2
Separate into ox and red and balance for elements except H and O
Cu → Cu2+ (Oxidation)
SO42- → SO2 (Reduction)
Assign oxidation #s and balance for electrons:
Cu → Cu2+ + 2eSO42- + 2e- → SO2
Balance charge with H+
Cu → Cu2+ + 2e4H+ + SO42- + 2e- → SO2
Balance O with H2O
Cu → Cu2+ + 2e2+
4H + SO4 + 2e- → SO2 + 2 H2O
Multiply be factor to eliminate 2e- (in this case 1)and add
Cu → Cu2+ + 2e4H+ + SO42- + 2e- → SO2 + 2 H2O
Cu + 4H+ + SO42- → Cu2+ + SO2 + 2 H2O
2 marks for correct answer
5. Balance the following in basic solution:
Zn + NO3- → Zn(OH)42- + NH3
Separate into ox and red and balance for elements except H and O
Zn → Zn(OH)42- + (Oxidation)
NO3- → NH3 (Reduction)
Assign oxidation #s and balance for electrons:
Zn → Zn(OH)42- + 2eNO3- + 8 e- → NH3
Balance charge with OH-
Balance O with H2O
4OH- + Zn → Zn(OH)42- + 2eNO3- + 8 e- → NH3 + 9 OH-
4OH- + Zn → Zn(OH)42- + 2e6 H2O + NO3- + 8 e- → NH3 + 9 OH-
Multiply be factor to eliminate e- and add
16OH- + 4Zn → 4 Zn(OH)42- + 8e6 H2O + NO3- + 8 e- → NH3 + 9 OH7OH- + 4Zn + 6 H2O + NO3-
→ 4 Zn(OH)42- + NH3 +
One mark for correct separation into ox and red. One additional mark for correct
answer.
Chem 020 Tutorial Test 2, Section 004
Wednesday October 17, 07/08
Name:
TA:
1. What is the pH of a solution prepared by adding 0.7 liters of 0.3 M KOH to 0.5 liters
of 0.3 M H2SO4? (Assume volumes are additive).
Moles H2SO4 = MV = 0.3 x 0.5 = 0.15; Moles H+ = 2 x 0.15 = 0.30
Moles KOH added = MKOH x VKOH = 0.3 x 0.7 = 0.21
Reaction is: H+ + OH- → H2O
After reaction moles H+ remaining = 0.30 – 0.21 = 0.09
[H+ ] = moles/ vol = .09/1.0 = .09 pH = 1.05
1 mark for recognizing H2SO4 is diprotic, 1 mark for the correct answer (accept pH = 1).
2. A bottle of concentrated HCl has a density of 1.70 g/ml. When 1.0 ml of this
concentrated solution is diluted to 1.5 liters the pH of the resulting solution is found to be
2.0. Find: a) the Molarity of the HCl in the original concentrated solution and b) the % by
mass of HCl in the original concentrated solution.
If pH = 2 then [HCl] = .01
The moles HCl in the diluted solution may be obtained from:
Moles HCl = Mdilute solution x Vdilute solution = 0.01 x 1.5 = 0.015
Therefore we know 1.0 ml of the original concentrated solution contained 0.015 moles
HCl and:
MHCl = moles/volume = 0.015/.001 = 15
Now we also know 1 ml of the original concentrated solution contains 0.015 moles HCl (
mass = moles x molar mass = 0.015 x 36.45 = 0.547 g and, from the density, we know I
ml of this solution weighs 1.70 g.
% HCl by mass = [mass HCl/mass solution] x 100 = 32.2 %
1mark for concentration HCl in the dilute solution, 1 additional mark for the rest.
3. A sample of an unknown triprotic acid weighing 9.8 grams is neutralized by 375 ml 0f
0.4 M Ca(OH)2. Find the molar mass of the acid.
If the acid is triprotic and base is dibasic then:
2H3A + 3 Ca(OH)2 → 3 H2O + 3 Ca2+ + 2 A31.5 moles of Ca(OH)2 → are required for each mole of the acid.
Moles Ca(OH)2 = MV = 0.4 x .375 = 0.15
Therefore moles acid = 0.15/1.5 = 0.1
And Mass = moles x Molar mass = 9.8 = 0.1 x MM, MM = 98
2 marks for correct answer.
4. Give the oxidation number of all the elements in each of the following compounds:
LiH
Li = 1+, H= -1 (one mark if both right)
KMnO4 K = 1+, Mn = 7+, O = -1 (one mark if all three right)
1.
Balance the following in basic solution:
F2 → F- + OF2
Separate into ox and red and balance for elements except H and O
F2 → OF2 (Oxidation) (note strictly O is oxidized the F- comes from the reduction
in the next step)
F2 → 2F- (Reduction)
Assign oxidation #s and balance for electrons:
F2 → OF2 + 2e2e- + F2 → 2FBalance charge with OH-
2OH- + F2
2e- +
Balance O with H2O
→ OF2 + 2eF2 → 2F-
2OH- + F2 → OF2 + 2e- + H2O
2e- + F2 → 2F-
Multiply by factor to eliminate e-, , in this case 1, and add
2OH- + 2F2
→ OF2 + 2F- +
H2O
One mark for correct separation into ox and red. One additional mark for correct
answer.(feel free to be generous, this one is tricky)
Chem 020 Tutorial Test 2, Section 004
Tuesday October 23, 07/08
Name:
TA:
1. What is the pH of a solution prepared by adding 0.7 liters of 0.3 M Ca(OH)2 to 0.5
liters of 0.3 M H2SO4? (Assume volumes are additive).
Moles H2SO4 = MV = 0.3 x 0.5 = 0.15; Moles H+ = 2 x 0.15 = 0.30
Moles Ca(OH)2 added = MCa(OH)2 x VCa(OH)2 = 0.3 x 0.7 = 0.21 ; Moles OH- = 2 x .21 =
0.42
Reaction is: H+ + OH- → H2O
After reaction moles OH- remaining = 0.42 – 0.30 = 0.12
[ OH- ] = moles/ vol = .12/1.2 = 0.10 pOH = 1.0 pH = 13
1 mark for recognizing H2SO4 is diprotic and reacts with one mole Ca(OH)2 , 1 mark for
the correct answer (accept pH = 1).
2. A bottle of concentrated HNO3 has a density of 1.80 g/ml. If the solution is 65% by
mass HNO3 what volume of this solution will have a pH of 1.5 when it is diluted to a
volume of 500 ml?
If the density is 1.80 g/ml then 1.0 liters weigh 1800 g and the mass of HNO3 = 0.65 x
1800 = 1170 g and moles HNO3 = mass/molar mass = 1170/63 = 18.57 = molarity HNO3
Now;
M(before dilution)V(before dilution) = M(after dilution)V(after dilution)
After dilution pH = 1.5 [HNO3] = .0316
18.57 x V(before dilution) = 0.0316 x 0.5
V(before dilution) = 8.5 x 10-4 liters = 0.85 ml
1mark for concentration acid in the dilute solution, 1 additional mark for the rest.
3. Jane Student has an upset stomach with pH = 0.90. Naturally she wishes to change the
pH to the more comfortable value of pH = 1.5. Assuming a volume of 1.5 liters, what
weight of Mg(OH)2 should she consume?
When pH = 0.9 moles H+ = MV = 10-0.9 x 1.5 = 0.188
When pH = 1.5 moles H+ = MV = 10-1.5 x 1.5 = 0.047
Moles H+ removed = 0.188 -0.047 = 0.141
Moles Mg(OH)2 required = 0.141/2 = 0.070
Mass Mg(OH)2 required = 0.070 x molar mass = 0.070 x 58 = 4.06 g
One mark for moles H+ removed, one for correct answer.
4. Give the oxidation number of all the elements in each of the following compounds:
H2O2
H = 1+, O = -1 (one mark if both right)
H3PO4 H = 1+, P = 5+, O = -2 (one mark if all three right)
5. Balance the following in basic solution and identify the oxidizing and reducing agents:
ZnO22- + Pb → Zn + PbO
Separate into ox and red and balance for elements except H and O
Pb → PbO (Oxidation)
ZnO22- → Zn (Reduction)
Assign oxidation #s and balance for electrons:
Pb → PbO + 2e2e- + ZnO22- → Zn
Balance charge with OH-
Balance O with H2O
2OH- + Pb → PbO + 2e2e- + ZnO22- → Zn + 4OH2OH- + Pb
2H2O + 2e- +
→ PbO +
2e- + H2O
ZnO22- → Zn + 4OH-
Multiply by factor to eliminate e-, , in this case 1, and add
H2O
+ Pb
+
ZnO22- → PbO +
Zn + 2OH-
Oxidizing agent is: ZnO22- , Reducing agent is Pb
One mark for correct balance, one mark for identifying oxidizing and reducing
agents.
Chem 020 Tutorial Test 2, Section 001
Tuesday October 23, 07/08
Name:
TA:
1. What is the pH of a solution prepared by adding 0.7 liters of 0.3 M Ca(OH)2 to 0.5
liters of 0.3 M H2SO4? (Assume volumes are additive).
Moles H2SO4 = MV = 0.3 x 0.5 = 0.15; Moles H+ = 2 x 0.15 = 0.30
Moles Ca(OH)2 added = MCa(OH)2 x VCa(OH)2 = 0.3 x 0.7 = 0.21 ; Moles OH- = 2 x .21 =
0.42
Reaction is: H+ + OH- → H2O
After reaction moles OH- remaining = 0.42 – 0.30 = 0.12
[ OH- ] = moles/ vol = .12/1.2 = 0.10 pOH = 1.0 pH = 13
1 mark for recognizing H2SO4 is diprotic and reacts with one mole Ca(OH)2 , 1 mark for
the correct answer (accept pH = 1).
2. A bottle of concentrated HNO3 has a density of 1.80 g/ml. If the solution is 65% by
mass HNO3 what volume of this solution will have a pH of 1.5 when it is diluted to a
volume of 500 ml?
If the density is 1.80 g/ml then 1.0 liters weigh 1800 g and the mass of HNO3 = 0.65 x
1800 = 1170 g and moles HNO3 = mass/molar mass = 1170/63 = 18.57 = molarity HNO3
Now;
M(before dilution)V(before dilution) = M(after dilution)V(after dilution)
After dilution pH = 1.5 [HNO3] = .0316
18.57 x V(before dilution) = 0.0316 x 0.5
V(before dilution) = 8.5 x 10-4 liters = 0.85 ml
1mark for concentration acid in the dilute solution, 1 additional mark for the rest.
3. Jane Student has an upset stomach with pH = 0.90. Naturally she wishes to change the
pH to the more comfortable value of pH = 1.5. Assuming a volume of 1.5 liters, what
weight of Mg(OH)2 should she consume?
When pH = 0.9 moles H+ = MV = 10-0.9 x 1.5 = 0.188
When pH = 1.5 moles H+ = MV = 10-1.5 x 1.5 = 0.047
Moles H+ removed = 0.188 -0.047 = 0.141
Moles Mg(OH)2 required = 0.141/2 = 0.070
Mass Mg(OH)2 required = 0.070 x molar mass = 0.070 x 58 = 4.06 g
One mark for moles H+ removed, one for correct answer.
4. Give the oxidation number of all the elements in each of the following compounds:
H = 1+, O = -1 (one mark if both right)
H2O2
H3PO4 H = 1+, P = 5+, O = -2 (one mark if all three right)
5. Balance the following in basic solution and identify the oxidizing and reducing agents:
ZnO22- + Pb → Zn + PbO
Separate into ox and red and balance for elements except H and O
Pb → PbO (Oxidation)
ZnO22- → Zn (Reduction)
Assign oxidation #s and balance for electrons:
Pb → PbO + 2e2e- + ZnO22- → Zn
Balance charge with OH-
Balance O with H2O
2OH- + Pb → PbO + 2e2e- + ZnO22- → Zn + 4OH2OH- + Pb
2H2O + 2e- +
→ PbO +
2e- + H2O
ZnO22- → Zn + 4OH-
Multiply by factor to eliminate e-, , in this case 1, and add
H2O
+ Pb
+
ZnO22- → PbO +
Zn + 2OH-
Oxidizing agent is: ZnO22- , Reducing agent is Pb
One mark for correct balance, one mark for identifying oxidizing and reducing
agents.
Chem 020 Tutorial Test 2, Section 004
Wednesday, October 24, 07/08
Name:
TA:
1. What is the pH of a solution prepared by adding 0.7 liters of 0.3 M Ca(OH)2 to 0.5
liters of 0.3 M H2SO4? (Assume volumes are additive).
Moles H2SO4 = MV = 0.3 x 0.5 = 0.15; Moles H+ = 2 x 0.15 = 0.30
Moles Ca(OH)2 added = MCa(OH)2 x VCa(OH)2 = 0.3 x 0.7 = 0.21 ; Moles OH- = 2 x .21 =
0.42
Reaction is: H+ + OH- → H2O
After reaction moles OH- remaining = 0.42 – 0.30 = 0.12
[ OH- ] = moles/ vol = .12/1.2 = 0.10 pOH = 1.0 pH = 13
1 mark for recognizing H2SO4 is diprotic and reacts with one mole Ca(OH)2 , 1 mark for
the correct answer (accept pH = 1).
2. A bottle of concentrated HNO3 has a density of 1.80 g/ml. If the solution is 65% by
mass HNO3 what volume of this solution will have a pH of 1.5 when it is diluted to a
volume of 500 ml?
If the density is 1.80 g/ml then 1.0 liters weigh 1800 g and the mass of HNO3 = 0.65 x
1800 = 1170 g and moles HNO3 = mass/molar mass = 1170/63 = 18.57 = molarity HNO3
Now;
M(before dilution)V(before dilution) = M(after dilution)V(after dilution)
After dilution pH = 1.5 [HNO3] = .0316
18.57 x V(before dilution) = 0.0316 x 0.5
V(before dilution) = 8.5 x 10-4 liters = 0.85 ml
1mark for concentration acid in the dilute solution, 1 additional mark for the rest.
3. Jane Student has an upset stomach with pH = 0.90. Naturally she wishes to change the
pH to the more comfortable value of pH = 1.5. Assuming a volume of 1.5 liters, what
weight of Mg(OH)2 should she consume?
When pH = 0.9 moles H+ = MV = 10-0.9 x 1.5 = 0.188
When pH = 1.5 moles H+ = MV = 10-1.5 x 1.5 = 0.047
Moles H+ removed = 0.188 -0.047 = 0.141
Moles Mg(OH)2 required = 0.141/2 = 0.070
Mass Mg(OH)2 required = 0.070 x molar mass = 0.070 x 58 = 4.06 g
One mark for moles H+ removed, one for correct answer.
4. Give the oxidation number of all the elements in each of the following compounds:
H = 1+, O = -1 (one mark if both right)
H2O2
H3PO4 H = 1+, P = 5+, O = -2 (one mark if all three right)
5. Balance the following in basic solution and identify the oxidizing and reducing agents:
ZnO22- + Pb → Zn + PbO
Separate into ox and red and balance for elements except H and O
Pb → PbO (Oxidation)
ZnO22- → Zn (Reduction)
Assign oxidation #s and balance for electrons:
Pb → PbO + 2e2e- + ZnO22- → Zn
Balance charge with OH-
Balance O with H2O
2OH- + Pb → PbO + 2e2e- + ZnO22- → Zn + 4OH2OH- + Pb
2H2O + 2e- +
→ PbO +
2e- + H2O
ZnO22- → Zn + 4OH-
Multiply by factor to eliminate e-, , in this case 1, and add
H2O
+ Pb
+
ZnO22- → PbO +
Zn + 2OH-
Oxidizing agent is: ZnO22- , Reducing agent is Pb
One mark for correct balance, one mark for identifying oxidizing and reducing
agents.