/10 KNOW
NAME:
SCH 4U ACID_BASE EQUILIBRIUM
/28 INQ
/38 TOTAL
PART ONE: KNOWLEDGE (10 Marks)
Multiple Choice (10 Marks)
Place the CAPITAL letter of the correct answer for each question in the table at the end of the questions.
1.
If a pH meter was placed in a 1.4 mol/L solution of nitric acid the reading would be which of the
following?
a. 1.4
d. 0.15
b. 14.15
e. 0.0
c. –0.15
2.
Which of the following salts could be combined with HC2H3O2 (oxalic acid) to form a buffer?
a. sodium oxalate
d. manganous cyanate
b. iron(III) gluconate
e. both c and d
c. sodium acetate
3.
When equal quantities of citric acid and lithium hydroxide are combined the solution is/has a
a. neutral
d. pH = 7
b. basic
e. both a and d
c. acidic
4.
If sodium formate was dissolved in distilled water, which of the following could be added to
make a functional buffer?
a. potassium formate
d. citric acid
b. formic acid
e. potassium citrate
c. formic hydroxide
5.
a.
b.
c.
d.
e.
6.
When 45 mL of 0.65 mol/L acetic acid is added to 65 mL of 0.45 mol/L sodium hydroxide the
resulting mixture is/has a(n)
a. neutral
d. pH < 7
b. basic
e. both c and d
c. acidic
7.
a.
b.
c.
d.
e.
8.
Kw is which of the following?
the equilibrium constant for water which is always 1.0  10-14
Ka  Kb for conjugate acid - base partners
the log [H2O]
both a and c
none of the above
A small amount of NaOH(aq) is added to this buffer system
HCHO2 + H2O <=====> H3O1+ + CHO21Which one of the following statements are true?
the pH drops only a little since the equilibrium shifts right
the pH rises only a little since the equilibrium shifts left
the pH drops only a little since the equilibrium shifts left
the pH rises only a little since the equilibrium shifts right
the pH does not change since the buffer uses up all the HCl(aq)
For sulfurous acid the Ka1 =
a. [SO32-][H1+]2 / [H2SO3]
b. [HSO42-][H1+] / [H2SO3]
c. [SO3 1-] [H1+]2/ [H2SO3]
d. [HSO31-][H1+] / [H2SO3]
e. [H2SO3] / [SO31-][H1+]2
If the Kb of a weak base is 2.9  10-8, the Ka of its conjugate acid partner must be which of the
following?
a. 2.9  10-8
d. 3.1  10-7
b. 6.46
e. 3.4  10-7
c. 7.54
Answer Table:
9.
1
2
3
4
5
6
7
8
9
C
A
B
B
B
B
D
D
E
PART TWO: INQUIRY (28 marks)
1. If the pH of a basic solution at 25oC is 12.56, what is the pOH; and the [H1+], [OH1-] in mol/L? (3
marks)
pOH =14-12.56 = 1.44
[H+]=10-12.56 = 2.8x10-13 M
[OH-]=10-1.44 =0.036 M
2. Calculate the pH and % ionization of a 0.45 M nitrous acid solution. Ka=7.2x10-4. (6 marks)
HNO2(aq) + H2O(l) ⇌ H3O+(aq) + NO2-(aq)
MR
I
C
E
1
0.45
-x
0.45-x
1
-----
1
0
x
x
1
0
x
x
Ka=[H3O+][NO2-]/[HNO2]
7.2x10-4 = x2/0.45-x
3.2 x10-4-7.2x10-4x-x2=0
a=-1, b= -7.2x10-4, c=3.2x10-4
x1= -0.018 x2 = 0.018
pH = 1.74
% ionization = [H3O+]/[HNO2] x 100%
= 4.0 %
3. What is the initial concentration of a weak base with Kb = 1.4  10-11 and pH = 8.75? (3 marks)
Base(aq) + H2O(l) ⇌ Hbase+(aq) + OH-(aq) Kb = 1.4  10-11
[Initial]-x
x
x
pOH=5.25
[OH-] = 10-5.25 = 5.6 x 10-6 = x
Kb= [Hbase+][OH-]/[base]
1.4x10-11=x2/I-x
1.4x10-11 I – 7.8x10-17 = (5.6x10-6)2
I =2.2M
4. A 28 mL of standardized 0.43 mol/L NaOH is titrated with 0.36 mol/L acetic acid
(Ka = 1.8 x 10-5). Calculate
a)
the volume of acetic acid required to reach equilibrium. (2 marks) 33 mL
b)
pH of the solution at equivalence. (5 marks) 9.02
HC2H3O2(aq) + NaOH(aq)  Na C2H3O2(aq) + H2O(l)
0.36
0.43
0.20
0.033
0.028
0.061
0.012
0.012
0.012
c
V
n
MR
I
C
E
C2H3O2-(aq) + H2O(l) ⇌ HC2H3O2(aq) + OH-(aq)
1
1
1
1
0.20
-0
0
-x
--x
x
0.20-x
-x
x
Kb= [HC2H3O2][OH-]/[ C2H3O2-]
1.0x10 /1.8x10-5 = x2/(0.20)
x=1.1x10-5
pOH=4.98
pH=9.02
-14
5. A buffer is made by mixing 0.80 mol of formic acid (HCHO2) with 0.90 mol of sodium formate
(NaCHO2) in 2.0 L of water. If the Ka value is 1.8x10-4, calculate
a) the pH of the initial buffer. (4 marks) 3.80
b) the pH of the solution after the addition of 0.10 mol of OH- ions. (5 marks) 3.63
MR
I
C
E
HCHO2 (aq) + H2O(l) ⇌ CHO2- (aq) + H3O+(aq)
1
1
1
1
0.40
-0.45
0
-x
--x
x
0.40-x
-0.45+x
x
Ka= [HC2H3O2][ H3O+]/[ HCHO2]
1.8x10-4 = x(0.45+x)/(0.40-x)
x=(1.8x10-4)(0.40)/0.45
x=1.6x10-3
pH=3.80
b) If we add hydroxide, the equilibrium will shift right increasing the formate concentration by 0.05
M to 0.50 M and decreasing the formic acid concentration by 0.05 M to 0.35 M.
Ka= [HC2H3O2][ H3O+]/[ HCHO2]
1.8x10-4 = x(0.50+x)/(0.35-x)
x=(1.8x10-4)(0.35)/0.50
x=1.3x10-4
pH=3.89