1
SCH4U FINAL EXAM REVIEW SOLUTIONS
KINETICS:
1. It would require the simultaneous collision of 27 molecules, which is an unlikely
event.
16 mol C6.O 2
 4.06 105 mol CO 2 L1s 1
2. 34 105 mol O 2 L1s 1
25 mol O 2
x
y
3. Rate = [O2] [NO] . To determine the exponents x and y we construct a ratio of
two rate laws using data from two different rows in our table
x
y
x
y
Rate1 k  O 2 1  NO1
3.29 106 k  0.00173  0.00347 
and


Rate2 k  O 2 x  NOy
6.55 106 k  0.00345x  0.00347 y
2
2
From this ratio we cancel the rate constants, k, and the [0.00347]y terms to get
x
3.29 106  0.00173
, Evaluating this ratio we get 0.4977 = (0.501)x. From

6.55 106  0.00345x
this we conclude that x must be 1.
To determine the value of y we substitute rows 1 and 3 of the table to get
x
y
3.29 106  0.00173 0.00347 
. We get 0.0628 = (0.250)y from which y = 2.

x
y
6
6.55 10
0.00173 0.0139
Therefore the rate law is Rate = k[O2][NO]2.
We can now calculate the rate constant by substituting the data from any row in
the table into the rate law. Using data from row 1 we get
3.29 x 10-6 mol L-1 s-1 = k(0.00173 mol L-1)(0.00347 mol L-1)2
k = 154 L2 mol-2 s-1.
4. The tool for solving this type of problem is the integrated rate equation for a firstorder process. One form of this equation is
ln[Q]o – ln[Q]t = kt. If we convert time to seconds we have all of the info
needed.
194 minutes = 1.164 x 104 s.
ln[Q]o – ln(8.54 x 10-6 mol L-1) = (3.64 x 10-4 s-1)(1.164 x 104 s)
ln[Q]o = 4.24 – 11.67 = -7.43
[Q]o = 5.9 x 10-4 mol L-1.
5.
O
O
N - - - -O- - - -N
O
E 1 1
k
6. ln 1  a    . All we need to do is convert the Celsius temperatures to
k2 R  T2 T1 
their Kelvin scale and insert the data to get
Ea
2.68 105
1 
 1
ln



5
1
1 
10.72 10
8.314 J mol K  313 K 293 K 


1.386  2.62 105 kJ 1 mol Ea
Ea  52.8 kJ mol1
The reaction rate
2
increased by a factor of four when the temperature was increased by 20oC. This is
the same as doubling every 10oC. Therefore, reactions that double their rates with
a ten degree rise in temperature have activstion energies of approximately 50 kJ.
7. Rate = k[(CH3)3CBr]
CHEMICAL EQUILIBRIUM
 NO2   K ;
c
2
3
 N 2O O2 
4
1. (a)
1
2
2. (c) HI(g )
3. K p 
(b)
C2 H 6   K
C2 H 4  H 2  c
H2 ( g )  12 I2 ( g ), Kc  0.142
4
PNO
2
PN22O PO22
4. (a) 3F2(g) + Br2(g)
2BrF3(g)
 BrF3 
3
 F2   Br2 
2
(b) K c 
5. K p  Kc ( RT )
ng
 (9.11080 )(0.0821 L atm mol 1 K 1  298 K) 1.0  3.7 1079
For H2(g) + O2(g)
H2O(g) .
The equation that we want Kp for has all of its coefficients one half of those given
with the value of Kc. That means that we must take the square root of the value of
Kc first and then convert to Kp.
(9.1 x 1080)1/2 = 3.2 x 1040.
Δngas = 1 mol – (1 mol + 0.5 mol) = -0.5 mol
n
K p  Kc ( RT ) g  (3.2 1040 )(0.0821 L atm mol1 K 1  298 K)0.5  6.47 1039
3
 Zn 2 
6. K c 
2
 Fe3 
7. We set up an equilibrium table to keep track of all the concentrations andtheir
changes. first we must accurately transfer data from the problem to the first brow
of the table. For the change line we insert x’s and the stoichiometric coefficients.
since the reaction goes in the forward direction, the 2x has a minus sign and the
other x values are positive. The third line is the sum of the first two and are the
terms inserted into the equilibrium law.
2HI
H2(g)
+
I2(g)
initial concentration 0.200
0.00500
0.00500
Change in
-2x
+x
+x
concentration
equilibrium
0.200 – 2x
0.00500 + x
0.00500 + x
concentrations
Setting up the equilibrium law we get:
3
 H 2  I2   0.00500  x0.00500  x  8.4 104
2
2
 HI
0.200  2 x
We can take the square root of both sides of the equation to get
0.00500  x   2.9 102
0.200  2 x 
Multiplying both sides by 0.200 – 2x results in
0.00500 + x = 5.8 x 10-3 – (5.8 x 10-2)x
x + 0.058x = 0.00580 – 0.00500
x = 7.6 x 10-4
Using this value of x we calculate that [HI] = 0.200 – 0.002 = 0.198 mol L-1
8.
2HBr(g)
H2(g)
+
Br2(g)
initial concentration 0.200
0.0
0.0500
Change in
-2x
+x
+x
concentration
equilibrium
0.200 – 2x
0.0 + x
0.0500 + x
concentrations
Setting up the equilibrium law we get:
 H2  Br2    0.0  x  0.0500  x   3.3 105
2
2
 0.200  2 x 
 HBr 
We will assume that x << 0.0500 and that 2x << 0.200 .
 x  0.0500   3.3 105
2
 0.200 
Solving this equation gives us a value x = 6.6 x 10-8 and [HBr] = 0.200 M.
9. (a) decrease (b) increase (c) no change (d) increase (e) no change
ACIDS AND BASES;
1. NH3 (acid) – NH4+ (base) ; NH3 (base) – NH2- (acid)
2. (a) pH = -log[0.25] = 0.60
(b) pOH = -log(.25) = 0.60; pH = 14 - pOH = 13.40
(c) pOH = -log(.0020) = 2.7; pH = 11.30
3. (a)
HCl(aq) + NaOH(aq)  H2O(l) + NaCl(aq) (overall)
+
H (aq) + OH-(aq)  H2O(l) (chemical reaction)
(b) nHCl = cV = (0.45 mol L-1)(0.0250 L) = 0.01125 mol
nNaOH = cV = (0.35 mol L-1)(0.0300 L) = 0.01050 mol. Since this is a 1:1
reaction [H+] = 0.1125 mol – 0.01050 mol = 7.5 x 10-4 mol. [H+] = n/V =
.0075/0.0750 L) = 0.01 mol L-1
pH = -log(0.01) = 2 (rounded!!).
4. Aniline has been identified as a weak base. Since it is the only solute, we will use
the Kb and start our problem by writing the chemical equation. Most weak bases
we see in this course are related to ammonia and we see that if the C6H5 in aniline
was replaced by a hydrogen atom it would have the formula of ammonia, NH3.
The advantage is that we know the equation for the reaction of ammonia with
water. NH3(aq) + H2O  NH4+(aq) + OH-(aq) .
We can use that as a template for our ionization reaction, or any other nitrogen
4
base. For the second part of the problem, to convert from the Kb to the pKb
pKb = -log Kb and to determine Ka involves KaKb = 1.0 x 10-14.
C6H5NH2(aq) + H2O C6H5NH3+(aq) + OH-(aq)
pKb = -log(1.5 x 10-10) = 9.82
1.0 1014
Ka =
 6.7 105 ; pKa = -log(6.7 x 10-5) = 4.17
10
1.5 10
5. First we identify that the category that this problem fits into is one where a weak
acid is the only solvent. second, we will write the ionization reaction for
propanoic acid and third, we write the equilibrium law.
C2H5COOH  H+ + C2H5COOThe equilibrium law is:
C2 H5COO   H  
x2
Ka 

initial concentration 
C 2 H5COOH
We will substitute the value of Ka (1.34 x 10-5) and the initial concentration of
propanoic acid (0.620 M) into the equation
x2
1.34  105 
0.620M 

x 2   0.620 M  1.34 105

x  2.88 103   H    C2 H 5COO  
Finally we calculate the remaining unionized propanoic acid by subtracting the
amount that does ionize.
[C2H5COOH] = 0.620 – 0.00288 = 0.617
pH = -log(2.88 x 10-3) = 2.54
6. When dissolved in water KNO2 undergoes the reaction
KNO2  K+(aq) + NO2-(aq)
If we add H+ to the NO2- we get HNO2 and recognize it as a weak acid. If we add
OH- to K+ we get KOH that is recognized as a strong base. Therefore K+ will not
affect the pH of the solution because it is a very weak conjugate acid, but the
NO2- will. It will act as a base (it must be a conjugate base since it came from an
acid HNO2). We will then write the reaction of NO2- with water as
NO2- + H2O
HNO2 + OHThe equilibrium law for this reaction is
 HNO2  OH  
Kb 
 NO2 
The initial concentration of NO2- is 0.100 M, and a small amount, x, reacts with
water. We then calculate that
[NO2-] = 0.100 – x and [HNO2] = [OH-] = x
To simplify the calculations we assume that x is very small so that [NO2-] =
0.100. We look up Ka = 7.1 x 10-4 for HNO2.
Kb = Kw/Ka = 1.4 x 10-11 = x2/0.100
5
x2 = (1.4 x 10-11)(0.100) = 1.4 x 10-12
x = 1.2 x 10-6 = [OH-]; pOH = 5.93; pH = 14.00 – pOH = 14.00 – 5.93 = 8.07
7. Yes, NH4+ reacts with water. pH = 4.93 (same method as #6).
8. CH3NH2(aq) + H2O  CH3NH3+(aq) + OH-(aq)
and


CH 3 NH 3  OH 
Kb  
CH3 NH 2 
CH3NH2(aq) + H2O  CH3NH3+(aq) + OH-(aq)
(2.5 x 10-3 – x)
x
x
xx
4.4  104 
2.5  103  x
x2 + (4.4 x 10-4)x – (1.1 x 10-7) = 0
b  b2  4ac 4.4 104  1.9 107  4(1)(1.1107 )
x

2a
2(1)
-4
-4
x = 1.8 x 10 and x = -6.2 x 10 . Only the positive values has any meaning so
[CH3NH2] = 2.5 x 10-3 – 1.8 x 10-4 = 2.3 x 10-3 M
[CH3NH3+] = [OH-] = 1.8 x 10-4 M
pOH = -log(1.8 x 10-4) = 3.74; pH = 14.00 - 3.74 = 10.26
9. The buffer solution contains both the weak acid CH3COOH and its conjugate base
CH3COO-. We can use either Ka or Kb, whichever is handy. In our tables we find
Ka = 1.8 x 10-6 for CH3COOH, so the simplest approach is to use the equation for
the ionization of the acid.
CH3COOH
H+ + CH3COO H   CH 3COO  
Ka 
 1.8 105
CH3COOH 
CH3COOH
0.090
-x
H+
+
CH3COO0.11
+x
initial concentration
0.0
Change in
+x
concentration
equilibrium
+x
0.090 – x  0.090
0.11 + x  0.11
concentrations
For buffer solutions the quantity x will be very small, so it is safe to make the
simplifying assumptions.
 x  0.11  x    x  0.11  1.8 105
Ka 
 0.090  x   0.090 
Solving for x gives us
 0.090 1.8 105  1.5 105 and [H+] = 1.5 x 10-5 M.
x
 0.11
pH = -log(1.5 x 10-5) = 4.82.
10. We begin with the chemical equation and the Kb expression.
NH3 + H2O
NH4+ + OH-
6
 NH 4  OH  
Kb  
 1.8  105
 NH3 
The key here is to use this Kb expression to calculate [NH4+].
We obtain [OH-] from the given pH: pOH = 14.00 - 9.10 = 4.90
[OH-] = 1.26 x 10-5 M.
Because Kb is small and the common ion NH4+ is present, the equilibrium
concentration of NH3 will essentially equal its initial concentration.
[NH3] = 0.25 M.
 NH3   1.8 105 (0.25)  0.36M
 NH 4   Kb
(1.26 105 )
OH  


Moles of NH4+ = (0.200 L)(0.36 mol L-1) = 0.071 mol
Mass of NH4Cl = nM = (0.071 mol)(53.5 g mol-1) = 3.82 g
11. C2H5COOH
H+ + C2H5COO-
OH   Kb   initial concnetration   7.4 1010  0.048  6.0 106 M .
The reaction of propanoic acid with KOH is
KOH + C2H5COOH
C2H5COO- + H2O
1. Determine the equivalence point volume by converting ml of propanoic acid to
mL of KOH using basic stoichiometric steps
25.00 mL C2 H 5COOH
0.0800 mol C2 H 5COOH
1 mol KOH
1 L KOH
 16.7 mL KOH
1 L C2 H5COOH
1 mol C2 H 5COOH 0.1200 mol KOH
The equivalence point is expected at 16.7 mL
2. The pH at the start of the titration is simply a solution containing a weak acid
and the hydrogen ion concentration is calculates, using the assumptions as
 H    K a  (initial concentration)  1.34 105  0.080  1.0 103 M
and the pH = -log(0.001) = 3.00
3. This part of the problem is actually a simple limiting reactant problem. Start by
calculating the moles of each substance we have.
nKOH = cV = (0.01000 L KOH)(0.1200 mol L-1 KOH) = 1.200 x 10-3 mol KOH
n C2 H5COOH  (0.02500 L)(0.0800 mol L-1) = 2.00 x 10-3 mol C2H5COOH
We have more moles of C2H5COOH than KOH so that all of the KOH is used up
forming an equal number of moles of C2H5COO- (1.200 x 10-3). Subtracting the
consumed KOH from the original moles of C2H5COOH leaves us with 0.80 x 10-3
mol C2H5COOH.
C2 H5COO   H  
1.200 103   H  
5
Ka 
 1.34 10 
0.80 103 
C 2 H5COOH 
Solving this for [H+] = 8.9 x 10-6M and the pH = -log(8.9 x 10-6) = 5.05
4. To calculate the pH at the equivalence point we need to know what the solution
contains. since we have reacted exactly equal moles of KOH and C2H5COOH
neither is present in any major amount. The solution is simply one that contains
the conjugate base of propanoic acid. We need to calculate Kb for this salt as
Kb = Kw/Ka = (1.0 x 10-14)/(1.34 x 10-5) = 7.5 x 10-10
7
The concentration of the conjugate base is calculated by dividing the moles of the
original acid by the total volume at the equivalence point.
[C2H5COO-] = (2.00 x 10-3 mol)(0.02500 L + 0.0167 L) 0.048M.
the hydroxide ion concentration can now be determined as
OH    Kb   initial concnetration   7.4 1010  0.048  6.0 106 M .
The pOH = 5.22 and the pH = 8.78 at the equivalence point.
5. To select an indicator we need to change its colour right at the equivalence
point. Therefore we look at the table of indicators and try to find one that has a
good colour change right around pH 8.78. In other words, the pKin should be
close to 8.78. There are several in this range including bromothymol blue and
phenolphthalein.
SOLUBILITY EQUILIBRIUM:
1.
CaSO4(s)
Ca2+(aq)
+
SO42-(aq)
initial concentration
0.0
0.0
Change in
+x
+x
concentration
equilibrium
+x
+x
concentrations
Ksp = 2.4 x 10-5 = [Ca2+][SO42-] = (x)(x) = x2
x = 4.9 x 10-3 M .
d = 1.0 g mL-1
4.9 103 mol CaSO4 1 mL solution 136 g CaSO4
0.67 g CaSO4
.


m=
.
1000 mL
1 g solution 1 mol CaSO4 1000 g solution
Since we want grams per 100 mL we divide the numerator and denominator by 10 to
get 0.067 g CaSO4 per 100 mL.
2. Same as above with common ion effect!!
CaSO4(s)
Ca2+(aq)
+
SO42-(aq)
initial concentration
0.0
0.14
Change in
+x
+x
concentration
equilibrium
+x
x + 0.14
concentrations
Ksp = 2.4 x 10-5 = [Ca2+][SO42-] = (x)(x + 0.14) = x2 + 0.14 x
If you recall that the solubility in the presence of a common ion is always less than in
distilled water, we can safely assume that (x + 0.14) = 0.14.
2.4 x 10-5 = (x)(0.14)
x = 1.7 x 10-4 M.
1.7 104 mol CaSO4 1 mL solution 136 g CaSO 4 0.023 g CaSO 4
.


and
1000 mL
1 g solution 1 mol CaSO4 1000 g solution
0.0023 g CaSO4 per 100 mL.
8
3. CaC2O4(s)
Ca2+ +
C2O42and
2+
2-9
Ksp = [Ca ][C2O4 ] = 2.3 x 10
Our final solution will have a total volume of 75.0 mL + 115 mL = 190 mL.
We can calculate the final concentration of calcium ions as
c1V1 = c2V2 and (4.5 x 10-5 M)(75.0 mL) = c2(190 mL)
c2 =[Ca2+] = 1.78 x 10-5 Mcalcium ions
For the oxalate ions we calculate
c1V1 = c2V2 and (1.3 x 10-6 M)(115.0 mL) = c2(190 mL)
c2 =[C2O42-] = 7.87 x 10-7 Moxalate ions
Now we can enter the two molarities into the Ksp expression to get
Q = (1.78 x 10-5 Mcalcium ion)(7.87 x 10-7Moxalate ion) = 1.4 x 10-11
We compare Q to Ksp and find that Q is much less than the value of Ksp and we
conclude that a precipitate will not form.
THAT’S ALL FOLKS!!!!!