SCH4U
June 2015
ANSWERS to EXAM REVIEW
1.
THERMOCHEMISTRY
The combustion of methanol is shown by the following equation:
a) Using the data below, find the heat of reaction for the combustion of methanol.
C(s) + O2(g)  CO2(g)
ΔH = -393 kJ (x2)
2H2(g) + O2(g)  2H2O(g)
ΔH = -484 kJ (x2)
2C(s) + 4H2(g) + O2(g)  2CH3OH(l) ΔH = -476 kJ (x (-1))
2C(s) + 2O2(g)  2CO2(g)
ΔH = -786 kJ
4H2(g) + 2O2(g)  4H2O(g)
ΔH = -968 kJ
2CH3OH(l) 2C(s) + 4H2(g) + O2(g)
ΔH = 476 kJ
2CH3OH(l) + 3O2(g)  2CO2(g) + 4H2O(g) ΔH = -1278
kJ
b) What is the molar enthalpy of combustion for methanol?
ΔH = -1278 kJ/2 mol = -639 kJ/mol
c) Express the energy changes for the target reaction in the form ΔH x =
ΔHcombustion CH3OH = -639 kJ/mol
d) Draw the potential energy diagram to illustrate the energy changes for the target reaction.
e) Write the target reaction that includes the heat term in the equation.
2CH3OH + 3O2  2CO2 + 4H2O + 1278 kJ
f) What mass of water could be heated from 20.00 C to 35.00 C by the burning of 10.0 g of methanol that is
held in a calorimeter with a heat capacity of 2.9 kJ/ oC?
qmethanol = qH2O
+
qcalorimeter
nΔH = mcΔT + CΔT
(0.312)(639000) = m(4.18)(15.0) + 2900(15.0)
m = 2480 g
n = 10.0/32.04 =0.312 mol
lΔHl = 639 kJ/mol = 639000 J/mol
Ccal = 2.9 kJ/oC = 2900 J/oC
g) Will entropy increase or decrease in the combustion of methanol? Will ΔS be positive or negative?
Entropy will increase because there will be an increase from 3 to 6 moles of gas. ΔS will be positive.
h) Calculate the change in entropy for the combustion of methanol.
ΔS = [2(214) + 4(189)] – [2(127) + 3(205)] = 315 J
i)
Consider your values of ΔH and ΔS. Will the combustion of methanol always be spontaneous? Explain.
ΔH is negative (i.e.the products are more stable than the reactants) and ΔS is positive (i.e. the products
are more randomly organized than the reactants). Both of these conditions favour a reaction occurring
therefore the combustion of methanol will always be spontaneous.
j)
Calculate the change in Gibb’s free energy for the combustion of methanol at 100 oC. Is the reaction
spontaneous or non-spontaneous at this temperature? Explain your choice.
ΔG = ΔH –TΔS
ΔS = 315 J = 0.315 kJ
= -1278 – (373)(0.315)
T = 273K + 100 = 373 K
= -1395 KJ
ΔG is negative therefore the reaction is spontaneous.
2. Consider the reaction:
NH4Cl(s)  NH3(g) + HCl(g)
What are the temperature requirements for the above reaction to work?
ΔH = [(-46) + (-92)] – [(-314)] = 176 kJ
ΔS = [193 + 187] – [96] = 284 J = 0.284 kJ
For the reaction to work, ΔG must be negative
OR
ΔH –TΔS < 0
176 – T(0.284) <0
-0.284T< -176
T> 619 K * remember to change the direction of the inequality when dividing by a negative value
Therefore the reaction will only work when the temperature is greater than 619.7 K or greater than 346.7 oC.
3. The molar enthalpy of combustion of butane is -2871 kJ/mol. What is the molar enthalpy of formation of
butane? HINT: first write out the balanced equation for the combustion reaction.
2C4H10(g) + 13O2(g)  8CO2(g) + 10H2O(l) ΔH = -5742 kJ
ΔH = [8(ΔHf CO2 ) + 10(ΔHf H2O )] – [2 ΔHf C4H10 + 13ΔHf O2]
-5742 = [8(-393) + 10(-286)] – [2 ΔHf C4H10 + 13(0)]
2 ΔHf C4H10 = -262
ΔHf C4H10 = -131 kJ
4.
a) Draw the heating curve for methanol, given: enthalpy of fusion = 3.2 kJ/mol; enthalpy of vaporization =
35.3 kJ/mol; heat capacity of liquid methanol = 2.5 J/g∙ oC; heat capacity gaseous methanol 1.6 J/g∙oC ;
the melting point of methanol = -97.6oC; boiling point of methanol = 64.7oC.
b) What amount of energy would be required to raise a 10.0 g sample of methanol from 0 oC to 100oC.
From 0oC to 64.7oC: (liquid)
Q = mcΔT = (10.0 g)(2.5 J/goC)(64.7-0) = 1620 J
Boiling:
10.0 g = 10.0/32.04 = 0.3121 mol
Q = (0.3121 mol)(35.3 kJ/mol) = 11.02 kJ = 11020 J
From 64.7oC to 100oC: (gas)
Q = mcΔT = (10.0 g)(1.6 J/goC)(100-64.7) = 565 J
Therefore the amount of energy to raise 10.0 g of methanol from 0oC to 100oC
= 1620 + 11020 + 565
= 13.2 kJ
RATES:
1. The following data was collected for the reaction: 2NO(g) + O2(g)  2NO2(g)
[O2] mol/L
[NO(g)] mol/L
0.0010
0.0040
0.0040
0.0010
0.0010
0.0030
Initial rate of formation of
NO2 (mol/L.s)
7.1
28.4
85.2
a) What is the rate law? Rate = k[O2][NO]
b) What is the value of the rate constant?
7.1 = k(0.0010)(0.0010)
K = 7.1 x 106 L/mol∙s
c) When the rate of formation of NO2 is 7.1 mol/L.s, what is the rate of disappearance of O2? 3.5
mol/L.s NO? 7.1 mol/L.s
d) Is it reasonable to suggest that this reaction proceeds as a 1-step reaction? No because according to
the rate law the rate-determining step involves 1 NO. Since there are 2 NO’s in the reaction, another
step is needed to introduce this second NO.
e) If the reaction proceeds as a 2-step mechanism, propose a mechanism that is consistent with the rate
law.
O2 + NO  NO3
(slow step)
NO3 + NO  2NO2 (fast step)
2. The following data was collected for the reaction: 2P + 3Q + R  T + 2U
Trial #
[P] (mol/L)
[Q] (mol/L)
1
2
3
1.0
1.0
1.0
1.0
2.0
2.0
[R]
(mol/L)
1.0
2.0
3.0
Rate of formation
of T (mol/L.min)
3.0
6.0
6.0
4
5
6
1.0
2.0
3.0
3.0
3.0
3.0
4.0
5.0
6.0
9.0
36
81
a) What is the rate law? Rate = k[P]2[Q][R]0 or Rate = k[P]2[Q]
b) What is the effect on the rate of i) doubling [P] rate x4 ii) tripling [Q] rate x3
iii) quadrupling [R] no effect
iv) increasing temperature? Rate increased
because ‘k’ will be bigger
c) What is the overall order of the reaction? 3rd order
d) Determine the rate constant. k = 3 L2/mol2∙s
e) What is the rate of formation of T when [P] = 4.0 M, [Q] = 3.0 M and [R] = 3.0 M? 144 mol/L.s
3. Draw a potential energy diagram for the reaction: A + B  C using the following information.
a) The mechanism consists of one elementary step.
b) The reaction is endothermic.
c) The activation energy for the reaction has a value 3x the enthalpy value of the reaction.
PE Diagram
PE
Ea
C
Δ
A+B
H
Reaction Progress
4. Consider the following reaction: N2(g) + 3H2(g)  2NH3(g). Suppose the reaction is first order in N2
and first order in H2,
a) Write the rate law. Rate = k[N2][H2]
b) Propose a probable mechanism?
N2 + H2  N2H2
(slow)
N2H2 + H2  NH3 + NH (fast)
NH + H2  NH3
(fast)
5. A reaction has the following mechanism:
2NO2  NO3 + NO (slow)
NO3 + CO  NO2 + CO2 (fast)
a) What is the rate law? Rate = k[NO2]2
b) What is the overall reaction? NO2 + CO  NO + CO2
c) The overall reaction has an enthalpy change of -228 kJ/mol. Draw the potential energy diagram for
the overall step.
PE Diagram
NO2 + CO
PE
ΔH = -228 kJ/mol
NO + CO2
Reaction Progress
d) Draw a potential energy diagram for the 2-step reaction if ∆H = -100 kJ for step one and ∆H for step
two is -128 kJ.
EQUILIBRIUM:
1. The dissociation of ammonia at 27oC has a Keq value of 2.63 x 10-9.
Heat + 2NH3(g) 
N2(g) + 3H2(g)
a) If 1.00 M ammonia is placed in a reaction vessel, calculate the equilibrium concentrations of N 2 and H2.
2NH3(g)  N2(g) + 3H2(g)
Keq = 2.63 x 10-9
[I] 1.00
0
0
[C] -2x
+x
+3x
[E] 1.00 – 2x
x
3x
** -2x is insignificant because 1.00 > 1000x Keq
Keq = [N2][H2]3
[NH3]2
2.63 x 10-9 = (x)(3x)3
(1.00)2
2.63 x 10-9 = 27x4
X = 0.00314
Therefore, [N2] = 0.00314 M and [H2] = 3(0.00314) = 0.00942 M
b) What is the percent yield of N2(g)?
If the reaction could go to completion, 1.00 mol of NH3 could produce 0.500 mol
N2. (2:1 reaction)
% yield = 0.00314
0.500
x 100% = 0.628%
c) Without adding more ammonia, what three things could be done to increase the yield of N 2?
Increase the temperature (reaction is endothermic)
Remove either N2 or H2
Decrease the pressure by increasing the volume of the container
d) What is the equilibrium constant for the reaction:
N2(g) + 3H2(g)  2NH3(g)?
K’ eq = [NH3]2
This is the reciprocal of the Keq in part a)
[N2][H2]3
= 1/Keq
= 1/(2.63 x 10-9)
= 3.80 x 109
2. 0.20 moles each of A, B and C are mixed into a 2.0 L container and allowed to react according to the following
equation: A(g) + B(g)  C(g)
[I]
[C]
[E]
Keq = 1.0 x 10-6. Calculate the equilibrium concentrations of each gas.
A(g) +
B(g)  C(g) Keq = 1.0 x 10-6
0.10
0.10
0.10
+x
+x
-x (see note below)
0.10+x 0.10+x 0.10-x
Q = [C]
= 0.10
= 10
[A][B] (0.10)(0.10)
Q>Keq therefore the reaction needs to proceed to the left.
Keq = [C]
[A][B]
= 0.10-x
(0.10+x)(0.10+x)
1.0x10-6 (0.010 + 0.20x +x2) = 0.10-x
1.0x10-8 + 2.0x10-7x + 1.0x10-6x2 = 0.10 – x
These values become insignificant when
like terms are collected
1.0x10 x + x - 0.10 = 0
X = -1 ± √(12 –4(1.0x10-6)(-0.10)
2(1.0x10-6)
X = 0.10
Therefore [A] = 0.10+0.10 = 0.20M
[B] = 0.20 M
[C] = 0.10-0.10 = 0 (This isn’t possible. We know that the value of x<0.10 but
-6 2
it has been rounded to 0.10 on the calculator. To get the non-zero value of C, we need to
resolve for C using our equilibrium values of A and B.
1.0x10-6 = [C]
(0.20)(0.20)
Therefore [C] = 4.0x10-8M
3. Consider the following equilibrium:
NH3(aq) + H2O(l)  NH4+(aq) + OH-(aq)
At equilibrium, the concentrations of NH3(aq), NH4+(aq) and OH-(aq) are 1.00M, 0.0010M and 0.018M
respectively. To 500 .0 mL of this equilibrium mixture, 0.050 mol of OH - are added. Calculate the new
equilibrium concentrations of NH4+ and OH-.
NH3(aq) + H2O(l)  NH4+(aq) + OH-(aq)
[I]
Stress
[C]
[E]
1.00
0.0010
+x
1.00+x
-x
0.0010-x
0.018
+ 0.10
-x
0.118-x
Keq = [NH4+][ OH-] = (0.0010)(0.018) = 1.8x10-5 (from initial equilibrium)
[NH3]
1.00
-5
1.8x10 = (0.0010-x)( 0.118-x) = 1.18x10-4 – 0.119x + x2
1.00+x
(insignificant since x<0.0010)
X – 0.119x + 1.0x10 = 0
X= 0.119 ± √(0.1192 –4(1)(1.0x10-4)
2
X = 8.46x10-4
Therefore [NH4+] = 0.0010 – 0.000846 = 1.54x10-4M
[OH-] = 0.118 – 0.000846 = 0.117M
2
-4
SOLUTIONS:
4. What is the Ksp of PbCl2 if, in a **saturated solution of lead (II) chloride, the [Cl-] is 0.032 mol/L?
[I]
[C]
[E]
PbCl2(s)  Pb2+(aq) + 2Cl-(aq)
0
0
+x
+2x
0.016
0.032
**saturated solution  at equilibrium
2X = 0.32
X = 0.016
Ksp = [Pb2+(aq)] [Cl-(aq)]2
= (0.016)(0.032)2
= 1.6 x 10-5
5. The solubility of calcium sulfate is **0.67 g/L. Calculate the Ksp for calcium sulfate.
[I]
CaSO4(s)  Ca2+(aq) + SO42-(aq)
0
0
[C]
[E]
(-x)
+ x
0.0049
+x
0.0049
Ksp = [Ca2+(aq)][ SO42-(aq)]
= (0.0049)(0.0049)
= 2.4 x 10-5
**Solubility in mol/L is “-x” in the ICE chart therefore need to change 0.67 g/L to mol/L
X = (0.67/136.14) = 0.0049 mol/L
6. The solubility of copper (II) hydroxide is **1.8 x 10-6 g/100 mL. Calculate the Ksp for Cu(OH)2.
Cu(OH)2(s) 
[I]
[C]
[E]
(-x)
Cu2+(aq)
0
+1.8 x 10-7
1.8 x 10-7
+
2OH-(aq)
0
+ 2(1.8 x 10-7)
3.6 x 10-7
Ksp = [Cu2+][OH-]2
= (1.8 x 10-7 )(3.6 x 10-7)2
= 2.3 x 10-20
**Solubility in mol/L is “-x” in the ICE chart therefore need to change 1.8 x 10-6 g/100mL to mol/L
X = (1.8 x 10-6/97.57)/0.100L = 1.8 x 10-7 mol/L
7. Calculate the calcium ion concentration in a saturated solution of calcium carbonate.
CaCO3(s)  Ca2+(aq) + CO32-(aq)
0
0
(-x)
+ x
+x
x
x
[I]
[C]
[E]
Ksp = [Ca2+(aq)][ CO32-(aq)]
8.7 x 10-9 = x2
x = 9.3 x 10-5
Therefore [Ca2+(aq)] = 9.3 x 10-5 mol/L
8. What is the solubility, in g/L, of calcium phosphate? Ksp for calcium phosphate is 1.2 x 10-26.
Ca3 (PO4)2
[I]
[C]
[E]
(-x)
(s)
 3Ca2+(aq) + 2PO43-(aq)
0
0
+ 3x
+ 2x
3x
2x
Ksp = [Ca2+(aq)]3 [ PO43-(aq)]2
1.2 x 10-26 = (3x)3(2x)2
108x5 = 1.2 x 10-26
X = 2.6 x 10-6
Therefore solubility is 2.6 x 10-6 mol/L OR (2.6 x 10-6 mol/L)(310.18 g/mol) = 8.1 x 10-4 g/L
9. Which is more soluble, calcium fluoride or silver chloride?
**Need to calculate the solubility for each and then compare “x”
[I]
[C]
[E]
CaF2(s)  Ca2+(aq) + 2F-(aq)
0
0
(-x)
+x
+2x
x
2x
Ksp = [Ca2+(aq)] [F-(aq)]2
4 x 10-11
= (x)(2x)2
-11
4 x 10
= 4x3
X = 2.2 x 10-4
Ag+(aq) +
0
+ x
x
AgCl(s) 
[I]
[C]
[E]
(-x)
Cl-(aq)
0
+x
x
Ksp = [Ag+(aq)][ Cl-(aq)]
1.6 x 10-10 = x2
x = 1.3 x 10-5
Since “x” represents the solubility, CaF2 is more soluble.
10. The concentration of Ag+ in a saturated solution of Ag2CrO4 is 1.5 x 10-4 mol/L. Calculate the Ksp for Ag2CrO4.
Ag2CrO4(s) 
[I]
[C]
[E]
(-x)
2Ag+(aq)
+
0
+1.5 x 10-4
1.5 x 10 -4
CrO42-(aq)
0
+ (1.5 x 10-4)/2
7.5 x 10-5
[Ag+] = 1.5 x 10-4
Ksp = (1.5 x 10-4)2 (7.5 x 10-5)
= 1.7 x 10-12
11. 1.0 x 10-3 moles of Ba2+ are added with 6.0 x 10-3 moles of SO42- to make a 1 L solution. What is the nature of
the solution?
BaSO4(s) 
Ba2+(aq) +
SO42-(aq)
Ksp = 1.5 x 10-9
1.0 x 10-3
[I]
6.0 x 10-3
Q= [Ba2+(aq)][ SO42-(aq)]
= (1.0 x 10-3)( 6.0 x 10-3)
= 6.0 x 10-6
Q > Ksp therefore the solution is a supersaturated solution.
12. If 2.00 mL of 0.200M NaOH are added to 1.00 L of 0.100M CaCl2, will a precipitate form?
Two products are formed: NaCl which is highly soluble and Ca(OH) 2 which has low solubility. If a
precipitate forms it will be calcium hydroxide.
Each solution added dilutes the other so concentrations need to be adjusted.
[NaOH] = [OH-]:
C1V1=C2V2
(0.200)(0.002) = C2(1.002)
C2 = 3.99 x 10-4M
[CaCl2 ] = [Ca2+ ]:
C1V1=C2V2
(0.100)(1.00) = C2(1.002)
C2 = 0.0998M
Ca(OH)2(s) 
[I]
Ca2+(aq)
0.0998
+
2OH-(aq)
3.99 x 10 -4
Q = [Ca2+][OH-]2
= (0.0998 )(3.99 x 10-4)2
= 1.6 x 10-8
Ksp = 1.3 x 10-6
Q < Ksp therefore a precipitate will not form
13. How much precipitate of BaCO3 will form when 20.0 mL of 0.10 M Ba(NO3)2 is added to 50.0 mL of 0.10 M
Na2CO3?
BaCO3(s) 
[I]
[C]
[E]
(+x)
[Ba(NO3)2] = [Ba2+]
C1V1=C2V2
(0.100)(20.0) = C2(70.0)
C2 = 2.86 x 10-2M
[Na2CO3 ] = [CO32- ]:
Ba2+(aq) +
2.86 x 10 -2
- x
2.86 x 10 -2 -x
CO32-(aq)
7.14 x 10-2
-x
Note: +x represents the amount of precipitate
-2
7.14 x 10 - x
C1V1=C2V2
(0.100)(50.0) = C2(70.0)
C2 = 7.14 x 10-2 M
Q>Ksp therefore a precipitate will form and the reaction will shift to the left
Ksp = [Ba2+(aq)][ CO32-(aq)]
1.6 x 10-9 = (2.86 x 10-2 –x)(7.14 x 10-2 – x)
1.6 x 10-9 = 0.00204 – 0.100x + x2 Note that when the two constants are collected, 1.6 x 10 -9
becomes
Insignificant; that is Ksp = 0
2
0 = 0.00204 – 0.100x + x
x = 2.86 x 10-2 or 7.14 x 10-2 – (too large)
Mass of precipitate formed = (2.86 x 10-2mol/L)(197.34 g/mol)(0.070L) = 0.395 g
14. Consider #10. When equilibrium is reached, what will a) [CO32-] be?
b) [Ba2+] be?
[CO32-] = 0.0714 – 0.0286 = 0.0428M
Note: [Ba2+] appears to be zero but isn’t actually zero but very close to it. This is because the K sp became
insignificant in our calculation. We need to bring it back in to find the barium ion concentration.
1.6 x 10-9 = [Ba2+](0.0428)
[Ba2+] = 3.74 x 10-8M
15. What mass of BaSO4 precipitate will form when 50 mL of 0.10M Na2SO4 is mixed with 50 mL of 0.20 M BaCl2?
BaSO4(s) 
[I]
[C]
[E]
(+x)
Ba2+(aq) +
0.10
- x
0.10 -x
SO42-(aq)
0.050
-x
0.050 - x
Note: +x represents the amount of precipitate
[BaCl2] = [Ba2+]
C1V1=C2V2
(0.20)(50) = C2(100)
C2 = 0.10M
[Na2SO4 ] = [SO42-]:
C1V1=C2V2
(0.10)(50) = C2(100)
C2 = 0.050 M
Q>Ksp therefore a precipitate will form and the reaction will shift to the left
Ksp = [Ba2+(aq)][SO42-(aq)]
1.5 x 10-9 = (0.10–x)(0.050 – x)
1.5 x 10-9 = 0.0050 – 0.150x + x2
Note that when the two constants are collected, 1.5 x 10-9 becomes
insignificant; that is Ksp = 0
0 = 0.0050– 0.150x + x2
x = 0.050 or 0.10 – (too large)
Mass of precipitate formed = (0.050mol/L)(233.39 g/mol)(0.100L) = 1.2 g
16. Calculate the molar solubility of BaSO4 in
a)
[E]
BaSO4(s) 
(-x)
Ba2+(aq) +
x
a) pure water
b) in 1.0 M SO42-.
SO42-(aq)
x
Note: -x represents the solubility
1.5 x 10-9 = x2
X = 3.9 x 10-5
The solubility in pure water is 3.9 x 10-5 M.
b)
[I]
[C]
[E]
Ba2+(aq) +
BaSO4(s) 
(-x)
0
+x
x
SO42-(aq)
1.0
+x
1.0 + x
Note: the sulfate will decrease the solubility of
barium sulfate therefore x < 3.9 x 10-5 .This
means
-x for sulfate will be insignificant
1.5 x 10-9 = 1.0x
X = 1.5 x 10-9
The solubility of barium sulfate in 1.0M sulfate is 1.5 x 10-9M.
17. Calculate the molar solubility of AgCl in a 1.0 L solution containing 10.0 g of CaCl2.
[I]
[C]
[E]
of
AgCl(s)  Ag+(aq) + Cl-(aq)
0
0.18
(-x)
+x
+x
x
0.18 + x This x will be insignificant. The common ion effect lowers the value
X and x< 10-5.
Cl- is the common ion
[Cl-] from CaCl2 = (10.0/111.08) x 2 = 0.18 (since 2 Cl- in CaCl2)
Ksp = [Ag+][Cl-]
1.6 x 10-10= 0.18x
X = 8.9 x 10-10
Therefore the solubility is 8.9 x 10-10 mol/L.
18. How many grams of CaCO3 will dissolve in 300 mL of 0.050 M Ca(NO3)2?
[I]
[C]
[E]
CaCO3(s)  Ca2+(aq) +
0.050
(-x)
+x
0.050 + x
CO32-(aq)
0
+x
X
Note: x will be insignificant; i.e. 𝑥 < √(8.7𝑥10−9)
Ksp = [Ca2+(aq)][ CO32-(aq)]
8.7 x 10-9 = 0.050x
X = 1.7 x 10-7
mass = (1.7 x 10-7)(100.09)(0.300) = 5.1 x 10-6g
19. a) Calculate the molar solubility of MgF2.
b) Calculate the solubility of MgF2 in g/100 mL.
c) How many grams of MgF2 will dissolve in a 350 mL solution of 0.25 M NaF?
d) If 50 mL of 0.50 M solution of CaCl2 are added to 250 mL of a saturated solution MgF 2, will a precipitate
form? Ksp for CaF2 is 3.9 x 10-11. Show your calculations.
[I]
[C]
[E]
MgF2(s)  Mg2+(aq) + 2F-(aq)
0
0
+x
+2x
x
2x
Ksp = [Mg2+(aq)] [F-(aq)]2
6.4 x 10-9
= (x)(2x)2
6.4 x 10-9
= 4x3
X = 1.2 x 10-3
a) Therefore the solubility is 1.2 x 10-3 mol/L
b) The solubility is (1.2 x 10-3)(62.31)(0.10) = 0.0075 g/L
[I]
[C]
[E]
MgF2(s)  Mg2+(aq) + 2F-(aq)
0
0.25
+x
+2x
x
0.25 + 2x Note: x<0.0012 since the common ion, F-, will reduce
the solubility, making 2x insignificant
Ksp = [Mg2+(aq)] [F-(aq)]2
6.4 x 10-9
= (x)(0.25)2
X = 1.0 x 10-7 = solubility in mol/L
c) Solubility in g/350mL = (1.0 x 10-7)(62.31)(0.350) = 2.2 x 10-6g
d)
CaF2(s)  Ca2+(aq) + 2F-(aq)
[I]
0.083 2.4 x 10 -3
Note: Initial [F-] comes from the saturated solution from part a)
Initial [Ca2+]:
C1V1=C2V2
(0.50)(50) = C2(300)
C2= 0.083
Q = [Ca2+][F-]2
= (0.083)(2.4 x 10-3) = 1.99 x 10-4
Q>Ksp therefore a precipitate will form.
ACIDS and BASES:
Ka and Kb
1.
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Calculate the Ka and percent ionization for 0.50M HNO2 if the pH is 1.7.
HNO2(aq) + H2O 
0.50
-0.020
0.48
H3O+(aq) +
0
+ 0.020
0.020
NO2-(aq)
0
+ 0.020
0.020
pH = 1.7
[H+] = 10-1.7
= 0.020M
Ka = (0.020)2 = 8.3 x 10-4
0.48
% ionization = 0.020 x 100% = 4.0%
0.50
2. Calculate the pH of 0.015M hydrazine, N2H4. Kb for hydrazine is 9.6 x 10-7.
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N2H4(aq) + H2O 
0.015
-x
0.015 – x
9.6 x 10-7 = x2
0.015
X = 1.2 x 10-4 = [OH-]
pOH = 3.9
pH = 10.1
N2H5+(aq) +
0
+x
x
OH-(aq)
0
+x
x
Kb = 9.6 x 10-7
3. The percent ionization of 0.10M methanoic acid is 4.2%. Calculate K a for methanoic acid.
HCO2H(aq) + H2O 
0.10
- 0.0042
0.10 – 0.0042
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E
H3O+(aq)
0
+0.0042
0.0042
HCO2-(aq)
+
4.2% of 0.10 = 0.0042
0
+0.0042
0.0042
Ka = (0.0042)2
(0.10 – 0.0042)
= 1.8 x 10-4
4. Calculate [H+], [H2C6H5O7 -], [HC6H5O7 2-], [C6H5O7 3-] in a 0.10M solution of citric acid, H3C6H5O7. Ka1= 7.1 x
10-4, Ka2= 1.7 x 10-5, Ka3= 6.3 x 10-6
H3C6H5O7(aq)
[E]
+
H2O 
H3O+(aq) + H2C6H5O7 –(aq)
0.10 – x
x
Ka1= 8.4 x 10-4
x
8.4 x 10-4 = x2
0.10 – x
8.4 x 10-4 (0.10 – x) = x2
X2 + 8.4 x 10-4x – 8.4 x 10-5 = 0
X = -8.4 x 10-4 + √(8.4 x 10-4)2 – 4(-8.4 x 10-5)
2
= 0.0087
Therefore [H+] = [H2C6H5O7 -] = 0.0087M
H2C6H5O7- (aq)
[E]
+
H2O 
0.0087 – x
H3O+(aq) + HC6H5O7 2–(aq)
0.0087 – x
Ka2= 1.8 x 10-5
x
1.8 x 10-5 = (0.0087)x
0.0087
X = 1.8 x 10-5
Therefore, [HC6H5O7 2-] = 1.8 x 10-5M
HC6H5O72- (aq)
E
1.8 x 10-5 – x
4.0 x 10-6 = (0.0087)x
1.7 x 10-5
-9
X = 7.9 x 10
+
H2O 
H3O+(aq) + C6H5O7 3–(aq)
0.0087+x
x
Ka1= 4.0 x 10-6
Therefore [C6H5O7 3-] = 7.9 x 10-9 M
Kw and Conjugate Acid-Base Pairs
5. What is the conjugate base of
a) HOCl?
OCl- Kb = Kw/Ka of HOCl = 1.0x10-14/3.5x10-8 = 2.9x10-7
b) CH3CO2H?
CH3CO2- Kb = Kw/Ka of CH3CO2H = 1.0x10-14/1.8x10-5 = 5.6x10-10
c) HNO2?
NO2- Kb = Kw/Ka of HNO2 = 1.0x10-14/4.0x10-4 = 2.5x10-11
d) HCO3 ?
CO32- Kb = Kw/Ka of HCO3- = 1.0x10-14/5.6x10-11 = 1.7x10-4
e) H2PO4-?
HPO42- Kb = Kw/Ka of H2PO4- = 1.0x10-14/6.2x10-8 = 1.6x10-7
Determine Kb for each.
6. What is the conjugate acid of:
a) N2H4?
N 2H 5+
Ka = Kw/Kb of N2H4 = 1.0x10-14/3.0x10-6 = 3.3x10-9
b) HCO3 ?
H2CO3 Ka = Ka1 = 4.3 x 10-7
c) PO43-?
H PO42 Ka = Ka3 = 4.8 x 10-13
Determine Ka for each.
Hydrolysis of Salts
7. Will solutions of the following salts be acidic, basic or neutral?
a) NH4CH3COO
b) Na2SO4
c) K2HPO4
d) NH4Cl
a)
e) NaCN
NH4+ (acidic?) CH3COO- (basic?)
Ka for NH4+ = 10-14/Kb for NH3 = 10-14/(1.8 x 10-5) = 5.6 x 10-10
Kb for CH3COO- = 10-14/Ka for CH3COOH = 10-14/(1.8 x 10-5) = 5.6 x 10-10
Ka = Kb Therefore the solution will be neutral
b) Na + (neutral) SO42- (basic?)
Kb for SO42- = 10-14/Ka for HSO4- = 10-14/1.0 x 10-2 = 10-12
Therefore basic
c) K+(neutral) HPO42- (acidic? basic?)
Ka for HPO42- = Ka2 = 4.2 x 10-13
Kb for HPO42- = 10-14/Ka for H2PO4- = 10-14/6.3 x 10-8 = 1.6 x 10-7
Kb > Ka therefore the solution will be basic
d) NH4+ (acidic) Cl-(neutral)
Ka for NH4+ = 10-14/Kb for NH3 = 10-14/(1.8 x 10-5) = 5.6 x 10-10
Therefore acidic
e) Na+ (neutral) CN- (basic?)
Kb for CN- = 10-14/Ka for HCN = 10-14/6.2 x 10-10 = 1.6 x 10-5
Therefore basic
8. A 0.250M solution of sodium ascorbate, Na2C6H6O6, has a pH of 8.65. Calculate Kb for the ascorbate ion.
C6H6O62- + H2O  HC6H6O6- +
OH-
pH = 8.65
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E
0.250
-4.47 x 10-6
0.250
0
0
-6
+4.47 x 10
+4.47 x 10-6
4.47 x 10-6
4.47 x 10-6
pOH = 5.35
[OH ] = 10-5.35 = 4.47 x 10-6
-
Kb = (4.47 x 10-6)2
0.250
= 7.98 x 10-11
9. Calculate the pH of 0.20M solution of NaCN
CN-(aq) + H2O  HCN(aq)
0.20 – x
x
[E]
1.6 x 10-5 = x2
0.20
X = 1.8 x 10-3
[OH] = 1.8 x 10-3 M
+
OH-(aq)
x
pOH = 2.7
Kb = 1.6 x 10-5
pH = 11.3
Titration
10.
a) If 250.0 mL of 0.300M Ca(OH)2 is required to titrate 600.0 mL of HNO2(aq), what is the concentration
of HNO2(aq)?
Ca(OH)2(aq) + 2HNO2(aq)  Ca(NO2)2(aq) + 2H2O
nCa(OH)2 = C x V = (0.300)(0.250) = 0.0750 mol
nHNO2 = 2(0.0750) = 0.150 mol
[HNO2]
= 0.150mol/0.600
L = 0.250 M
b) What is the pH of the equivalence point?
The NO2- ions in the salt that is produced will act as a base and affect the equivalence point.
Kb for NO2- = Kw/Ka of HNO2 = 1.0 x 10-14 / 4.0 x 10-4 = 2.5 x 10-11
nCa(NO2)2 = 0.0750 mol
[Ca(NO2)2] = 0.0750 mol/0.850 L = 0.0882M
[NO2-] = 2(0.0882) = 0.176M
[E]
NO2-(aq)
+
0.176 – x
H2O  HNO2(aq) + OH-(aq)
x
x
2.5 x 10-11 = x2/0.176
X = 2.1 x 10-6
pOH = 5.7
pH = 8.3
Therefore the pH at the equivalence point will be 8.3
c) What would be an effective acid-base indicator to use?
11. A 8.24 g sample of a solid organic acid is dissolved in water. It is found that 47.3 mL of 0.100 M NaOH are
needed to titrate the acidic solution. Assuming that there is only one hydrogen in the acidic solid,
determine the molar mass of the solid acid. What would the molar mass be if the acid has two hydrogens?
HA + NaOH -- NaA + H2O
nNaOH = cxv = (0.0473)(0.100) = 0.00473 mol
nHA = 0.00473 mol
mm = m/n = 8.24 g/0.00473 mol = 1742 g/mol
If acid is diprotic:
H2A + 2NaOH -- Na2A + 2H2O
nNaOH = cxv = (0.0473)(0.100) = 0.00473 mol
nHA = 0.00473/2 = 0.00237 mol
mm = 8.24g/0.00237 mol = 3477 g/mol
12. 30.0 mL of 0.10M HCl is added to 30.0mL of 0.15M NaOH. What is the resulting pH of the solution?
Note: both the acid and base are strong.
nHCl = (0.10mol/L)(0.030L) = 0.0030 mol
nNaOH = (0.15mol/L)(0.030L) = 0.0045 mol
n NaOH remaining un-neutralized = 0.0045 – 0.0030 = 0.0015mol
[NaOH] = 0.0015mol/0.060L = 0.025M
pOH = 1.6
pH = 12.4
Buffers
13. How many grams of KF must be dissolved in 250 mL of 0.50M HF so that the pH of the solution is 3.0?
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HF(aq) + H2O
0.50
-10-3
0.50 -10-3

H3O+(aq)
0
+10-3
10-3
+
F-(aq)
x
+10-3
x + 10-3
Ka = 6.6 x 10-4
pH = 3.0 therefore [H+] = 10-3M
6.6 x 10-4 = (10-3)(x + 10-3)
0.50 – 10-3
X = 0.33
[F-] = 0.33M = [KF]
M = (0.33mol/L)(0.250L)(58.1 g/mol) = 4.8 g
14. 6.0 g of sodium benzoate and 6.0 g of benzoic acid are added to water to make a 500 mL solution. Calculate
the pH of the resulting solution.
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C6H5O2H + H2O 
0.098
-x
0.098 – x
C6H5O2-(aq)
0.083
+x
0.083 + x
+ H3O+(aq)
0
+x
x
Ka = 6.3 x 10-5
n C6H5O2H = 6.0/122.12 = 0.0491
[C6H5O2H] = 0.0491/0.500 = 0.098
n C6H5O2 = 6.0/144.1 = 0.0416
[C6H5O2-] = 0.0416/0.500 = 0.083
6.3 x 10-5 = 0.083x
0.098
X = 7.4 x 10-5
pH = 4.1
15. Create a buffer system that has a pH of 5.5.
ATOMIC THEORY and BONDING
1. Compare the following terms:
a) Orbit, orbital
b) Bohr’s model of the atom, quantum mechanical model
c) Polar molecule, non-polar molecule
d) Electron configuration, orbital diagram
e) P orbital, s orbital, hybrid orbital
f) London forces, dipole-dipole forces
g) Dipole-dipole forces , hydrogen bonding
h) Covalent bond, VanderWaals forces
i) Sigma bond, pi bond (AP)
2. Draw the electron configuration for:
a) Manganese
1s2 2s2 2p6 3s2 3p6 4s2 3d5
b) antimony (short form)
[Kr] 5s2 4d10 5p3
c) the phosphorus ion
1s2 2s2 2p6 3s2 3p6
3. Draw the orbital diagram for:
a) Titanium (short form)
b) Fe2+
c) Carbon in its promoted state
4. Draw the orbital diagram for:
a) Potassium’s 19th electron
b) Valence electrons of aluminum
5. Draw the bonding orbitals and label the orbitals for:
a) PBr3
b) SiCl4
c) CH3F
6. Draw structural diagrams for:
a) O3
b) HNO3
c) HCN
7. Draw and name 3-D shape for:
a) HCN
b) CO32c) PBr3
d) SiCl4
e) PBr5
f) SF4
g) SF6
h) BrCl3
i) XeF2
j) XeF4
8. Determine whether each of the compounds in #7 is polar.
10. Which compounds in #7 required electrons to be promoted to form a hybrid orbital?
Name the hybrid orbital formed.
a) HCN
sp hybrids
2b) CO3
sp2 hybrids (AP)
c) PBr3
no hybridization necessary
d) SiCl4
sp3 hybrids
e) PBr5
sp3d hybrids
f) SF4
p3d hybrids
g) SF6
sp3d hybrids
h) BrCl3
p2d hybrids
i) XeF2
pd hybrids
j) XeF4
p2d2 hybrids
11. Determine the predominant intermolecular force acting on each substance in #7.
a) HCN polar therefore dipole-dipole and London forces
b) CO32- non-polar London forces (and ionic)
c) PBr3
polar therefore dipole-dipole and London forces
d) SiCl4 non-polar therefore London forces
e) PBr5 non-polar therefore London forces
f) SF4
polar therefore dipole-dipole and London forces
g) SF6 non-polar therefore London forces
h) BrCl3 polar therefore dipole-dipole (minor since only very slightly polar) and London
forces
i) XeF2 non-polar therefore London forces
j) XeF4 non- polar therefore London forces
12. Consider the following pairs. Predict which has the higher boiling point. State the main
reason why.
a) XeF2 , XeF4 both non-polar therefore only London forces at work. The number of
electrons in the molecule is the greatest factor that impacts strength of London forces.
Since XeF4 has more electrons it will have the highest boiling point.
b) H2, F2 both non-polar therefore only London forces at work. The number of electrons in
the molecule is the greatest factor that impacts strength of London forces. Since F2 has
more electrons it will have the highest boiling point.
c) CH2Cl2, CCl4 CH2Cl2 is polar therefore has dipole-dipole and London forces at work. CCl4
has only London forces since it is non-polar. However this is difficult to determine which
will have the higher boiling point since one could argue that CH2Cl2 has dipole-dipole and CCl4
does not it would cause it to have the highest. However, CCl4 has greater London forces
(greater number of electrons) therefore one could argue that it has the greater boiling
point. In actual fact, the boiling point of CH2Cl2 is 39.6oC and CCl4 is 76.7oC so the greater
London forces wins.
d) H2O, H2S
Both are polar. H2S has dipole –dipole and London forces at work; H2O has Hydrogen
bonding and London forces at work. The hydrogen bonding in water far outweighs the
greater London forces in H2S and therefore water has the highest boiling point.
 Try also the self-quizzes for chapters 3 and 4 and Unit 2