TOPIC 4 : ACID-BASE EQUILIBRIA
1
List three strong acids and three strong bases.
[6]
ANY THREE
Strong acids: HCl, HBr, HI, HNO3, HClO4, HClO3, H2SO4
Strong bases: NaOH, KOH, LiOH, Ca(OH)2, Ba(OH)2, Sr(OH)2
4(a)
2
Classify the following compounds as strong acid, strong base, weak acid or
weak base.
[8]
H2SO4
CsOH
H2SO3
CH3CH2NH2
HIO3
NH3
Mg(OH)2
HBrO
Strong acid
Strong base
Weak acid
Weak base
H2SO4
CsOH
H2SO3
CH3CH2NH2
HIO3
Mg(OH)2
HBrO
NH3
4(a)
3
Explain in terms of dissociation, why sodium hydroxide is a strong base while
ammonia is a weak base.
[2]
NaOH is a strong base because of it dissociates completely in water,
while NH3 is a weak base because it dissociates partially in water.
NaOH yields high concentration of OH- ions.
4(c)
4
The concentrations of hydronium ion, [H3O+] for acid A, B and C are 7.0 x 103
M, 3.5 x 10-2 M and 9.5 x 10-5 M respectively. Arrange these 3 acids
according to the increase in acidity. Explain your answer and relate it to acid
dissociation constant, Ka.
[3]
C<A<B
As concentration of hydronium ion increases, acidity increases. High
concentration of hydronium ion comes from high degree of acid
dissociation. Ka = [A-][H3O+]/[HA]. As [H3O+] increases, Ka
increases.
4(d)
5
Given that the Ka values for phenylacetic acid, propanoic acid and pyruvic
acid are 4.9 x 10-5, 1.3 x 10-5 and 2.8 x 10-3 respectively. Arrange these
acids according to increasing acid strength. Explain.
[3]
Propanoic acid<phenylacetic acid<pyruvic acid
The higher the Ka value, the higher the degree of dissociation in
water. This leads to higher concentration of hydronium ion. Hence,
more acidic.
4(d)
6
In each equation, state the conjugate acid-base pairs.
[4]
(a) NH3 (aq) + HNO3 (aq) ⇌ NH4+ (aq) + NO3- (aq)
NH3 + HNO3 ⇌ NH4++ NO3B
A
CA
CB
Conjugate acid-base pairs: NH3/NH4+ and HNO3/ NO3-
(b) O2- (aq) + H2O (l) ⇌ OH- (aq) + OH- (aq)
O2- + H2O ⇌ OH- + OHB
A
CA
CB
Conjugate acid-base pairs: O2/OH- and H2O/OH4(b)
7
In each equation, state the conjugate acid-base pairs.
[4]
(a) NH4+ (aq) + BrO3- (aq) ⇌ NH3 (aq) + HBrO3 (aq)
NH4+ + BrO3- ⇌ NH3 + HBrO3
A
B
CB
CA
Conjugate acid-base pairs: NH4+/NH3 and BrO3-/ HBrO3
(b) HClO (aq) + H2O (l) ⇌ ClO- (aq) + H3O+ (aq)
HClO (aq) + H2O (l) ⇌ ClO- (aq) + H3O+ (aq)
A
B
CB
CA
Conjugate acid-base pairs: HClO/ClO- and H3O+/H2O
4(b)
8
Write the Ka or Kb expression for the dissociation of the following acid or
base.
[2]
(a) HClO (aq) + H2O (l) ⇌ ClO- (aq) + H3O+ (aq)
Ka = [ClO-][ H3O+]/[ HClO]
(b) NH3 (aq) + H2O (l) ⇌ NH4+ (aq) + OH- (aq)
Kb = [NH4+][OH-]/[NH3]
4(e)
9
Write the equation for dissociation of the following acid or base. State the
conjugate acid-base pairs. Write the Ka (or Kb) expressions.
[12]
(a) HOCN
HOCN (aq) + H2O (l) ⇌ OCN- (aq) + H3O+ (aq)
Ka = [OCN-][ H3O+]/[ HOCN]
Conjugate acid-base pairs: HOCN/OCN- and H3O+/H2O
(b) CH3NH2
CH3NH2 (aq) + H2O (l) ⇌ CH3NH3+ (aq) + OH- (aq)
Kb = [CH3NH3+][OH-]/[CH3NH2]
Conjugate acid-base pairs: CH3NH3+/CH3NH2 and H2O/OH(c) H2S
H2S has 2 ionizable protons, hence, it dissociates twice in water.
1st dissociation
H2S (aq) + H2O (l) ⇌ HS- (aq) + H3O+ (aq)
Ka1 = [HS-][H3O+]/[H2S]
Conjugate acid-base pairs: H2S/HS- and H3O+/H2O
2nd dissociation
HS- (aq) + H2O (l) ⇌ S2- (aq) + H3O+ (aq)
Ka2 = [S2-][H3O+]/[ HS-]
Conjugate acid-base pairs: HS-/S2- and H3O+/H2O
4(e)
10
Determine if the following solutions are acidic or basic. Explain.
(a)
[6]
A solution with [OH-] = 9.0 x 10-11
A solution is basic when [OH-] > 1 x 10-7. Thus, this solution is
acidic since [OH-] < 1 x 10-7.
OR
pOH = -log (9.0 x 10-11)
pH
= 10.04
= 12 – 10.04 = 1.96
Hence, acidic.
(b)
A solution with [H3O+] = 1.0 x 10-10
A solution is acidic when [H3O+] > 1 x 10-7. Thus, this solution
is basic since [H3O+] < 1 x 10-7.
OR
pH
= -log (1.0 x 10-10)
= 10
Hence, basic.
(c)
A solution with pOH of 2.5
A solution with pOH of 2.5 has pH of 11.5. This solution is basic
since pH > 7.0
11
Based on the table below, arrange the compounds in order of increasing pH.
Show all the calculations involved.
[4]
pH of X
Compound
Concentration, M
X
[H3O+]=1.80 x 10-12
Y
[OH-]=2.40 x 10-4
Z
[H3O+]=7.20 x 10-6
= -log [H3O+]
= -log 1.8 x 10-12 = 11.75
pH of Y
= pKw – pOH
= 14 – (-log 2.4 x 10-4) = 10.38
pH of Z
= -log [H3O+]
= -log 7.2 x 10-6= 5.14
Order: Z<Y<X
4(g)
12
Calculate the [OH-] of a solution, given that the [H3O+] is 2.8 x 10-3 M.
Determine whether the solution is acidic, basic or neutral.
[2]
[H3O+]=Kw/[OH-]= 1.0 x 10-14/2.8 x 10-3 M
= 3.57 x 10-12 M
[OH-] > [H3O+] ; solution is basic
4(f)
13
Calculate the [H3O+] of a solution, given that the [OH-] is 2.8 x 10-3 M.
Determine whether the solution is acidic, basic or neutral.
[2]
[OH-] = Kw/[H3O+]
= 1.0 x 10-14/5.7 x 10-11 M
= 1.75 x 10-4 M
[OH-] > [H3O+] ; solution is basic
4(f)
14
The pH of phenylamine is 8.35. Calculate the [H3O ], [OH ] and pOH of the
solution at 25ºC.
[3]
+
pH
-
=-log [H3O+]
[H3O+]=10-pH = 10-8.35 = 4.47 x 10-9 M
[OH-] = Kw/[H3O+]
= 1.0 x 10-14/4.47 x 10-9 M
= 2.24 x 10-6 M
pOH = pKw – pH = 14 – 8.35 = 5.65
4(g)
15
The [H3O+] of CH3COOH is 7.2 x 10-6 M. Calculate the [OH-], pH and pOH at
25ºC.
[3]
[OH-] = Kw/[H3O+]
= 1.0 x 10-14/ 7.2 x 10-6 M
= 1.39 x 10-9 M
pH
=-log [H3O+] = 5.14
pOH = pKw – pH = 8.86
4(g)
16
Calculate the pH and pOH 1.0M of HNO3 and 1.0M LiOH of at 25ºC.
Determine the similarity between these 2 compounds.
[5]
Similarity: Both compounds are strong acids (or base), so they
dissociate completely in water.
(a)
1M HNO3 makes [H3O+] = 1M ; pH = - log [H3O+] = 0
[OH-] = (1 x 10-14)/1 = 1 x 10-14
;
pOH= - log [OH-] = 14
OR
(b)
pOH = 14 – pH = 14
1M LiOH makes [OH-] = 1M
; pOH = - log [OH-] = 0
[H3O+] = 1 x 10-14/1 = 1 x 10-14
;
pH = - log [H3O+] = 14
OR
pH = 14 – pOH = 14
4(g)
17
Calculate the Ka and pKa of 0.010 M HA, given that the [A-] is 1.4 x 10-8 M.
[3]
[], M
HA
H2O
A-
H3O+
I
0.01
-
0
0
C
-x
-
+x
+x
E
0.01-x
-
x
x
From the table, x = [A-] = [H3O+] = 1.4 x 10-8
Ka
= [A-][ H3O+]/[HA]
= x2/(0.01-x)
= (1.4 x 10-8)/(0.01-(1.4 x 10-8))
= 1.96 x 10-14
pKa = -log Ka
=13.71
4(h)
18
Calculate the Ka and pKa of 0.80 M HOCN, given that the pH is 1.48.
[3]
[H3O+]=10-pH = 10-1.48 = 0.033 M
[], M
HOCN
H2O
OCN-
H3O+
I
0.80
-
0
0
C
-x
-
+x
+x
E
0.80-x
-
x
x
From the table, x = [OCN-] = [H3O+] = 0.033 M
Ka
= [OCN-][ H3O+]/[HOCN]
= x2/(0.80-x)
=(0.033)/(0.80 -(0.033))
= 1.42 x 10-3
pKa
= -log Ka
= 2.85
4(h)
19
Given that the Kb value of 0.5 M of methylamine, CH3NH2 is 4.4 x 10-4.
Calculate the concentration of hydroxide ions, OH-.
[2]
[], M
CH3NH2
H2O
CH3NH3+
OH-
I
0.5
-
0
0
C
-x
-
+x
+x
E
0.5-x
-
x
x
Kb
= [CH3NH3+][OH-]/[CH3NH2]
= x2/(0.5-x)
0.5/Kb >> 400 ; x is neglected, we assume that [CH3NH2]eq ≈
[CH3NH2]i
Hence, the Kb value can be simplified to x2/(0.5)
x2/(0.5)
x
= 4.4 x 10-4
=[OH-] = 0.015 M
4(h)
20
1.0 M of CH3COOH has a Kb value of 5.56 x 10-10. Calculate the concentration
of hydronium ion, [H3O+] for this solution.
[3]
(M)
CH3COOH
H2O
CH3COO-
H3O+
I
1.0
-
0
0
C
-x
-
+x
+x
E
1.0-x
-
x
x
Ka=x2/1.0-x
Ka= Kw/Kb = (1 x 10-14)/5.56 x 10-10
= 1.8 x 10-5
Ka
= [CH3COO-][ H3O+]/[ CH3COOH]
= x2/1.0-x
1.0/Ka >>400, x is neglected, we assume that [CH3COOH]eq ≈
[CH3COOH]i
x2/1.0
x
= 1.8 x 10-5
= 4.24 x 10-3 M = [H3O+]
4(h)
21
Predict and explain whether the following salt is acidic, basic or neutral.
[4]
(a) CH3COONa
Basic salt. CH3COO- is basic because it is derived from a weak
acid while K+ is neutral because it is derived from a strong
base. CH3COO- will hydrolyze to form OH(b) NH4Cl
Acidic salt. NH4+ is acidic because it is derived from a
weak base while Cl- is neutral because it is derived
from a strong acid. NH4+ will hydrolyze to form H3O+
4(i)
22
Predict whether lithium bromide, LiBr is an acidic, basic or neutral salt.
Explain.
[2]
Neutral salt. Both Li+ and Br- ions are neutral because they are
derived from strong base and strong acid respectively.
4(i)
23
For each of the following, write the relevant acid-base equilibrium equation
and predict if pH increases or decreases with each change. Explain your
prediction.
[4]
(a)
LiCN is added to HCN
(b)
NH4Br is added to NH3
(a)
HCN (aq) + H2O (l) ⇌ CN- (aq) + H3O+ (aq)
Addition of CN- will shift equilibrium to the left (reactant side)
and
produce more HCN which leads to lesser H3O+. Thus, pH
will increase as acidity is reduced.
(b)
NH3 (aq) + H2O (l) ⇌ NH4+ (aq) + OH- (aq)
Addition of NH4+ will shift equilibrium to the left (reactant side)
and produce more NH3 which leads to lesser OH-. Thus, pH will
decrease as basicity is reduced.
4(j)
24
For each of the following, predict if pH increases or decreases with each
change. Explain your prediction.
[4]
(a)
CsOH is added to NaOH
(b)
KCl is added to HCl
(a)
Both CsOH and NaOH will dissociate completely in water. Thus,
addition of CsOH only adds more OH-, increasing the pH of
solution.
(b)
KCl is a neutral salt. Thus, the addition of KCl has no effect
whatsoever on the pH of solution.
4(j)
25
Define buffer.
[2]
A solution (mixture) of weak acid and its conjugate base or weak
base and its conjugate acid that can resist pH changes when a small
amount of strong acid or base is added into it
4(k)
26
A 1.0 L buffer solution is made up of 0.80 M HClO2 and 0.40 M NaClO2. [Ka =
1.1 x 10-2]
[6]
(a)
Calculate the initial pH of the buffer solution
(b)
Calculate the pH if 0.010 mol of KOH is added into the buffer solution
(c)
Calculate the pH if 0.010 mol of HCl is added into the buffer solution
[Assume the volume does not change after each addition]
(a) pH
= pKa + log [A-]/[HA]
= pKa + log [ClO2-]/[HClO2]
= -log(1.1 x 10-2) + log (0.4)/(0.8)
= 1.66
(b) 0.010 mol of OH- will react with 0.010 mol HClO2 to form 0.010
mol ClO2pH
= -log(1.1 x 10-2) + log (0.40+0.01)/(0.80-0.01)
= 1.67
(c) 0.010 mol of H3O+ will react with 0.01 mol ClO2- to form 0.01 mol
HClO2
pH
= -log(1.1 x 10-2) + log (0.40-0.01)/(0.80+0.01)
= 1.64
4(m)
27
Calculate the initial pH of a 1.0 L buffer solution that is made up of 0.10 M
NH3 and 0.50 M NH4Cl. [Kb = 1.75 x 10-5]
[3]
Ka
=(1 x 10-14)/(1.75 x 10-5)
= 5.71 x 10-10
pH = pKa + log [A-]/[HA]
= - log (5.71 x 10-10) + log (0.10/0.50)
= 8.54
4(m)
28
30 mL of 0.20M propanoic acid, CH3CH2COOH is titrated with 0.10M NaOH.
(Ka of CH3CH2COOH = 1.3 x 10-5)
(a)
Predict whether the pH at equivalence point will be acidic, basic or
neutral.
(b)
Calculate the initial pH
(c)
Calculate the pH after 10 mL of NaOH is added
(d)
Calculate the pH after 60 mL of NaOH is added
(e)
Calculate the pH after 70 mL of NaOH is added
(f)
Identify a suitable indicator that can be used in this titration
(g)
Sketch the titration curve
(a)
Basic. Due to the presence of CH3CH2COO- ion which will
hydrolyze to form OH-.
(b)
[], M
CH3CH2COOH
H2O
CH3CH2COO-
H3O+
I
0.20
-
0
0
C
-x
-
+x
+x
F
0.20 - x
-
x
X
0.20/Ka >>>400. Neglect x, assume [CH3CH2COOH]initial ≈ [CH3CH2COOH]eq
Hence,
Ka= x2/0.20
x2/0.20
= 1.3 x 10-5
x
= 1.612 x 10-3 M = [H3O+]
pH
= -log [H3O+]
= 2.79
(c)
n CH3CH2COOH initial
= (0.20)(0.03)
= 6 x 10-3 mol
n OH- added
= (0.10)(0.01)
= 1 x 10-3 mol
n, mol
CH3CH2COOH
OH-
CH3CH2COO-
H2O
I
6 x 10-3
1 x 10-3
0
0
C
-1 x 10-3
-1 x 10-3
+1 x 10-3
0
F
5 x 10-3
0
1 x 10-3
0
pH
= pKa + log [CH3CH2COO-]/[CH3CH2COOH]
= -log (1.3 x 10-5) + log (1 x 10-3/0.04 L)/(5 x 10-3/0.04 L)
= 4.19
(d)
n CH3CH2COOH initial
= 6 x 10-3 mol
n OH- added
= 6 x 10-3 mol
This is at the equivalence point whereby 6 x 10-3 mol of OH- reacts
with 6 x 10-3 mol of CH3CH2COOH to form 6 x 10-3 mol of CH3CH2COO-.
CH3CH2COO- will hydrolyze in water as such:
n, mol
I
CH3CH2COO6 x 10-3
H2O
CH3CH2COOH
OH-
-
0
0
C
-x
-
+x
+x
F
6 x 10-3 - x
-
x
x
Kb = Kw/Ka = 1.08 x 10-9
[Initial]/Kb >>> 400. We neglect x and assume [CH3CH2COO-]initial ≈
[CH3CH2COO-]eq
[CH3CH2COO-]
= 6 x 10-3 mol / 0.09 L
= 0.067 M
x2/0.067
x
= 1.08 x 10-9
= Kb
= 8.51 x 10-6 M
= [OH-]
pOH = -log OH-
= 5.07
pH = pKw – pOH = 8.93
(e) This is beyond the equivalence point whereby an excess of 0.010
L x 0.10M = 1 x 10-3 mol OH- is present.
[OH-] = (1 x 10-3 mol)/(0.030+0.070)L = 0.01 M
pH = 14 – pOH = 14 – (- log 0.01) = 12
(f) Phenolphtalein indicator
(g) Sketch of pH-volume curve
29
30 mL of 0.15M NH3 is titrated with 0.10 M HBr. [Kb (NH3) = 1.8 x 10-5]
(a)
Predict whether pH at the equivalence point will be acidic, basic or
neutral.
(b)
Determine a suitable indicator for this reaction.
(c)
Calculate the initial pH.
(d)
Calculate the pH after 10 mL of HBr is added
(e)
Calculate the pH after 45 mL of HBr is added
(f)
Calculate the pH after 60 mL of HBr is added
(g)
Sketch the titration pH curve
(a)
Acidic pH due to the presence of NH4+ ion which will hydrolyze
to form H3O+
(b)
Methyl red
(c)
Initial pH
[], M
NH3
H2O
NH4+
OH-
I
0.15
-
0
0
C
-x
-
+x
+x
F
0.15 - x
-
x
x
0.15 M/Kb >> 400, we can neglect x, we assume [NH3]eq≈[NH3]
so the Kb expression can be simplified into
Kb = x2/0.15
;
x = [OH-] = 1.64 x 10-3 M ;
pH = 14 - pOH = 14 – (-log 1.64 x 10-3) = 11.21
(d)
10 mL HBr
n NH3 initial = 4.5 x 10-3 mol
n HBr or H3O+ added = 1 x 10-3 mol
n, mol
NH3
H3O+
I
0.030Lx0.15M
0.010Lx0.10M
= 4.5 x 10-3
= 1 x 10-3
NH4+
H2O
0
-
C
- 1 x 10-3
- 1 x 10-3
+ 1 x 10-3
-
F
3.5 x 10-3
0
1 x 10-3
-
From the table, you can see that this is prior to equivalence point.
Also, base and its conjugate acid are present.
pH
= pKa + log [A-]/[HA]
= - log [(1 x 10-14)/(1.8 x 10-5)] + log [(3.5 x 10-3
mol/0.04L)/(1 x 10-3 mol/0.04L)]
= 9.80
(e)
45 mL HBr
This is at the equivalence point whereby 4.5 x 10-3 mol of HBr
reacts with 4.5 x 10-3 mol of NH3 to form 4.5 x 10-3 mol of
NH4Br. NH4Br will hydrolyze in water as such:
(mol)
NH4+
I
4.5 x 10-3
H2O
NH3
H3O+
-
0
0
C
-x
-
+x
+x
F
4.5 x 10-3 - x
-
x
x
Ka
= Kw/Kb = 5.56 x 10-10
= x2/(4.5 x 10-3-x) ≈ x2/(4.5 x 10-3) ;
can be simplified since 0.15/Ka >>400
x = [H3O+] = 1.58 x 10-6
pH = - log 1.58 x 10
-6
(f)
;
= 5.80
60 mL HBr
This is beyond the equivalence point whereby an excess of
0.015 L x 0.10M = 1.5 x 10-3 mol HBr is present.
[H3O+] = (1.5 x 10-3 mol)/(0.045+0.060)L = 0.0143 M
pH = - log 0.0143 = 1.84
(g)
Sketch of pH-volume curve