Ch. 15 & 16 Review
Everything except Polyprotics & Lewis
Acids/Bases!!
Practice 1
Write the equilibrium-constant expressions for
the following processes:
a)
H2(g) + I2(g)  2HI(g)
b)
Cd4+(aq) + 4Br-(aq)  CdBr4(aq)
c)
P4 (s) + 5 O2 (g) ↔ P4O10 (s)
Practice 2

For the UNBALANCED reaction:
NH3 (aq) ↔ N2 (g) + H2 (g)
1.
Write an equilibrium expression
2.
Calculate the value of K at 127˚C for:
 [NH3] = 3.1x10-2M
 [N2] = 8.5x10-1 M
 [H2] = 3.1x10-3 M
3.
Calculate the value of K with the above
concentrations for the following reaction:
1/2 N2 + 3/2 H2 ↔ NH3
Practice 3

A mixture of hydrogen and nitrogen in a
reaction vessel is allowed to attain
equilibrium at 472°C. The equilibrium
mixture of gases was analyzed and found
to contain 7.38 atm H2 , 2.46 atm N2 , and
0.166 atm NH3. From these data, calculate
the equilibrium constant Kp for the reaction
Practice 4
Sulfur trioxide decomposes at high
temperature in a sealed container:
Initially, the vessel is charged at 1000 K
with SO3(g) at a partial pressure of 0.500
atm. At equilibrium the SO3 partial
pressure is 0.200 atm. Calculate the value
of Kp at 1000 K.
Practice 5
At 448°C the equilibrium constant Kc for the
reaction
is 50.5. Predict in which direction the
reaction will proceed to reach equilibrium
at 448°C if we start with 2.0  10–2 mol of
HI, 1.0  10–2 mol of H2, and 3.0  10–2
mol of I2 in a 2.00-L container.
Practice 6
Consider the equilibrium
In which direction will the equilibrium shift when
(a) N2O4 is added
(b) NO2 is removed
(c) the total pressure is increased by addition of
N2(g)
(d) the volume is increased
(e) the temperature is decreased
7.) Determine if the following is acidic,
basic, or neutral.

Cu(NO3)2
Answer



Ions = Cu+2, NO3Cu+2 + H2O ↔ Cu(OH)2 + H+
weak base
**acidic (due to H+)
NO3- + H2O ↔ HNO3 + OHstrong acid
**neutral
(will dissociate, so H+ and OH- will form water)
 Salt
= Acidic
8.) Determine if the following is acidic,
basic, or neutral.

KClO4
Answer


Ions = K+, ClO4K+ + H2O ↔ KOH + H+
strong base
**neutral
(will dissociate, so H+ and OH- will form water)

ClO4- + H2O ↔ HClO4 + OHstrong acid
**neutral
(will dissociate, so H+ and OH- will form water)

Salt = neutral
9.) Determine if the following is acidic,
basic, or neutral.

NaH2PO4
Answer


Ions = Na+, H2PO4Na+ + H2O ↔ NaOH + H+
strong base
**neutral
(will dissociate, so H+ and OH- will form water)

H2PO4- + H2O ↔ H3PO4 + OHweak acid
**basic (due to OH-)
 Salt
= basic
10.) Determine if the following is acidic,
basic, or neutral.

LiF
Answer


Ions = Li+, FLi+ + H2O ↔ LiOH + H+
strong base
**neutral
(will dissociate, so H+ and OH- will form water)

F- + H2O ↔ HF + OHweak acid
 Salt
= basic
**basic (due to OH-)
11) Determine if the following is acidic,
basic, or neutral.

(NH4)2CO3

Extra info:


Kb (of NH3) = 1.8x10-5
ka (of HCO3-) = 5.6x10-11
Answer



Ions = NH4+, CO3-2
NH4+ + H2O ↔ NH3 + H3O+
weak base
**acidic (due to H O )
CO3- + H2O ↔ HCO3- + OHweak acid
**basic (due to OH-)
3
+
**must compare Kb vs. Ka to decide pH of salt
Kb (of NH3) = 1.8x10-5
Ka of NH4+ = 5.6x10-10
ka (of HCO3-) = 5.6x10-11
kb of CO3-2 = 1.8x10-4
 Salt = basic since Kb is greater than ka
12.) Complete the following table:
pH
[H+]
pOH
[OH–]
5.4 x 10–4
7.8 x 10-10
10.75
5.00
Acidic, basic,
or neutral?
Answers:


ROW1
pH = 3.27
pOH = 10.73
[OH–] = 1.9 x 10–11
acidic (since pH < 7)
ROW 2
pH = 4.89
[H+] = 1.3 x 10–5
pOH = 9.11
acidic (since pH < 7)


ROW 3
[H+] = 1.8 x 10-11
pOH = 3.25
[OH–] = 5.6 x 10–4
basic (since pH > 7)
ROW 4
pH = 9.00
[H+] = 1.0 x 10–9
[OH–] = 1.0 x 10–5
basic (since pH > 7)
Practice 13

Calculate the pH of a 0.0430 M HNO3
solution.
Answer

Since HNO3 is a strong acid, the nitric acid
solution will be 100% ionized.
Thus [H+] = [NO3–] = 0.0430 M.

The pH = - log [0.0430] = 1.37

Practice 14

Calculate the pH of a 0.020 M Ba(OH)2 (aq)
solution.
Answer

Since Ba(OH)2 is a strong base it is 100%
ionized. Note that ionization gives 2 OH– ions
for each mole of Ba(OH)2.
Thus [OH–] = 2 x 0.020 M = 0.040 M
pOH = -log[0.040] = 1.40

pH = 14 – 1.40 = 12.60


Practice 15

Calculate the pH of a 0.250 M HC2H3O2
solution. Ka(HC2H3O2) = 1.8 x 10-5.
Answer
HC2H3O2 
Balanced Equation
Answer:
H+
+
C2H3O2–
Initial Concentration (M)
0.250
0
0
Change (M)
-x
x
x
Equilibrium Concentration (M)
0.250 - x
x
x
(a)

2
2
[H ][C2 H3 O2 ]
x
x
5
Ka 
 1.8 10 

[HC2 H3 O2 ]
0.250- x 0.250
Thus x2 = 4.5 x 10-6; x = 2.12 x 10-3 = [H+].
pH = 2.67.
Practice 16

Calculate the pH of a 0.600 M solution of
methylamine CH3NH2. Kb = 4.4 x 10–4.
Answer
Answer:
Since CH3NH2 is a weak base, the balanced equation for the reaction is CH3NH2 + H2O  CH3NH3+ + OH–.
Balanced Equation
CH3NH2
Initial Concentration (M)
0.600
Change (M)
-x
Equilibrium Concentration (M)
0.600 - x
H 2
___
___
___
CH3NH3+ +
OH–
0
0
x
x
x
x

[BH ][OH] [CH3NH3 ][OH ]
x2
x2
4
Kb 



 4.4 10
[B]
[CH 3NH 2]
0.600 x 0.600
Thus x = 1.62 x 10-2 = [OH–], and pOH = 1.79.
The pH = 12.21.
Practice 17

The pH of a 0.10 M solution of a weak base
is 9.67. What is the Kb of the base?
Answer
Answer:
The balanced equation for a weak base B is given in Eq(10). The equilibrium table required is given below.
Balanced Equation
B
Initial Concentration (M)
0.10
Change (M)
-x
Equilibrium Concentration (M)
0.10 - x
H 2
___
___
___
BH+
+
OH–
0
0
x
x
x
x
At equilibrium, [OH–] = [BH+] = x. Use the pH to calculate the [OH–] at equilibrium
(which is the value of x).
Here pOH = 14.00 – pH = 14.00 – 9.67 = 4.33. Thus :

pOH
[OH ]  10
4.33
 10
5
 4.68 10  x
[BH ][OH ]
x2
x2
(4.68105 ) 2
Kb 



 2.2 108
[B]
0.10  x 0.10
0.10
Practice 18
Use the following acidity constants to help answer
the questions below:




1.
2.
3.
4.
Ka(HC2H3O2) = 1.8 x 10– 5
Ka(HCN) = 4.9 x 10–10
Ka(HCOOH) = 1.7 x 10-4
Which of the three acids is the weakest?
Which of the following bases is the strongest:
C2H3O2-, CN- , or HCOO- ?
What is the pKa of HCN?
What is the Kb for CN- ?
Answer




(1) smallest k value = HCN
(2) strong base = weakest acid = lowest k
value = HCN (acid) = CN- (base)
(3) pKa = -log Ka = -log (4.9x10-10) = 9.31
(4) Ka x Kb = 1x10-14
kb = 1x10-14 = 2.0 x 10-5
4.9x10-10
19.) Predict whether an aqueous solution would be
acidic, basic or neutral?
1.sodium nitrate
2.ammonium iodide
3.sodium bicarbonate
4.ammonium cyanide
5.sodium hypochlorite
6.potassium acetate
Predict whether an aqueous solution would be acidic,
basic or neutral? ANSWERS
1.sodium nitrate = NaNO3 = neutral
2.ammonium iodide = NH4I = acidic
3.sodium bicarbonate = NaHCO3 = basic
4.ammonium cyanide = NH4CN = basic (Kb ↑)
kb (for NH3) vs. Ka (for HCN)
1.8x10-5
vs. 4.9x10-10
5.sodium hypochlorite = NaClO = basic
6.potassium acetate = KC2H3O2 = basic
Practice 20 (from ch. 15 study questions)
Suppose that 0.50 moles of hydrogen gas, 0.50
moles of iodine gas, and 0.75 moles of hydrogen
iodide gas are introduced into a 2.0 Liter vessel
and the system is allowed to reach equilibrium.
H2(g) + I2(g) ↔ 2 HI(g)
Calculate the concentrations of all three substances
at equilibrium. At the temperature of the
experiment, Kc equals 2.0 x 10-2.

Practice 21 (from ch. 15 study questions)
Nitrosyl chloride NOCl decomposes to nitric oxide and chlorine
when heated:
2 NOCl(g) ↔ 2 NO(g) + Cl2(g)
At 600K, the equilibrium constant Kp is 0.060. In a vessel at
600K, there is a mixture of all three gases. The partial pressure
of NOCl is 675 torr, the partial pressure of NO is 43 torr and the
partial pressure of chlorine is 23 torr.
a.
What is the value of the reaction quotient?
b.
Is the mixture at equilibrium?
c.
In which direction will the system move to reach
equilibrium?
d.
When the system reaches equilibrium, what will be the
partial pressures of the components in the system? (just
set up the problem, do not solve all the way!!)
