Worksheet 20 – Polyprotic Acids and Salt Solutions
Ka
strong acid
1.3 x 10-2
7.1 x 10-4
6.8 x 10-4
1.8 x 10-5
4.5 x 10-7
9 x 10-8
5.6 x 10-10
6.2 x 10-10
4.7 x 10-11
1 x 10-17
negligible acidity
Acid
HNO3, HI, HCl, etc
HSO4HNO2
HF
CH3COOH
H2CO3
H2S
NH4+
HCN
HCO3HSLi2O, NaOH
Base
NO3-, I-, Cl-, etc
SO42NO2FCH3COOHCO3HSNH3
CNCO32S2O2-, OH-
Kb
negligible basicity
7.7 x 10-13
1.4 x 10-11
1.5 x 10-11
5.6 x 10-10
2.3 x 10-8
1.1 x 10-7
1.8 x 10-5
1.6 x 10-5
1.8 x 10-4
1 x 103
strong base
Salts are ionic compounds which dissociate in water to produce ions. They are formed
in the neutralization reaction between acids and bases. Depending on the nature of the
acids and bases (strong or weak), the solutions of the salts will be acidic, basic or
neutral.
1.
Decide which of the following salts will form acidic, basic or neutral solutions
when dissolved in water. (Hint: look at the acids and bases that formed them)
For example: KCH3COO was formed in the reaction between KOH and
CH3COOH
KOH is a strong base. Its conjugate acid, K+, has negligible acidity and will
leave the pH at 7.00, a neutral solution.
CH3COOH is a weak acid, making its conjugate base, CH3COO- a relatively
strong base. It will produce a basic solution.
a) KF
b) KCN
basic
basic
c) NaNO3
d) RbI
neutral
neutral
e) NH4NO3
f) Na2CO3
acidic
basic
2.
Rank the following 0.1 M aqueous salt solutions in order of increasing pH.
(Hint: write out the reactions of the salts + water)
a) KNO3
neutral
K2SO4
basic (Kb = 7.7x10-13)
K2S
basic (Kb = 1x103)
strongest base
pH rank: KNO3 < K2SO4 < K2S
b) NH4NO3
acidic
Ka = 5.6x10-10
NaHSO4
acidic
Ka = 1.3x10-2
strongest acid
NaHCO3
basic
Kb = 2.3x10-8
Na2CO3
basic
Kb = 1.8x10-4
strongest base
pH rank: NaHSO4 < NH4NO3 < NaHCO3 < Na2CO3
3.
In an experiment, it is found that the pHs of three salts, KX, KY and KZ are 7.0,
9.0 and 11.0. Arrange the acids, HX, HY and HZ in order of increasing acid
strength.
The strongest acid would have the weakest conjugate base. Weaker bases
(at equal concentrations) will have lower pH values (they are less basic).
The base strengths must be X- < Y- < ZThe conjugate acids will have the reverse trend: HZ < HY < HX
4.
Calculate the pH of a 0.1 M NaCN solution.
Na+ does not affect the pH of the solution
CN- is a weak base with Kb = 1.6x10-5
CN- + H2O ' HCN + OH-
Initial
Change
Equil.
[CN-]
[HCN]
[OH-]
0.10
-x
0.10 – x
0
+x
x
0
+x
x
K b = 1.6 × 10 −5 =
[HCN ][OH − ] ≈
[CN ]
−
x ≈ 1.3 × 10 −3
pOH = − log(1.3 × 10 −3 ) = 2.9
pH = 14 − 2.9 = 11.1
x2
0.10
5.
Suppose that 50.00 mL of 0.10M CH3COOH is combined with 50.00 mL of 0.10M
NaOH. What is the pH of the resulting system?
The equation for this reaction is:
CH3COOH + OH- Æ CH3COO- + H2O
a)
(neutralization reaction)
How many moles of CH3COO- are formed? (assume the reaction goes to
completion)
0.10 mol CH 3 COOH
mol CH 3 COOH = 0.050 L ×
= 0.0050 mol
1L solution
0.10 mol OH mol OH = 0.050 L ×
= 0.0050 mol
1L solution
moles
CH3COOH
OH-
CH3COO-
Initial
Change
Equil.
0.0050
-0.0050
0
0.0050
-0.0050
0
0
+0.0050
0.0050
This reaction would form 0.0050 moles of CH3COOb)
What is the final volume of the solution?
Final volume is the sum of the components: 50 mL + 50 mL = 100 mL
c)
What is the concentration of CH3COO-?
[CH3COO-] = 0.0050 mol / 0.100 L = 0.050 M
d)
Characterize CH3COO- (i.e. weak/strong acid/base).
weak base
e)
Write the reaction of the product (CH3COO-) and water.
CH3COO- + H2O ' CH3COOH + OH-
f)
Set up an ICE table for the reaction and calculate the pH of the solution.
[CH3COOH]
[OH-]
[CH3COO-]
Initial
0.050
0
0
Change
-x
+x
+x
Equil.
0.050 – x
x
x
K b = 5.6 × 10 −10 =
[CH 3COOH ][OH − ] ≈
[CH COO ]
−
3
x ≈ 5.3 × 10 −6
pOH = − log(5.3 × 10 −6 ) = 5.3
pH = 14 − 5.3 = 8.7
x2
0.050
6.
A 0.10 M solution of KOC6H5 has a pH of 11.40. Calculate the Ka value for
HOC6H5.
[OC6H5-]
[HOC6H5]
[OH-]
Initial
0.10
0
0
Change
-x
+x
+x
Equil.
0.10 – x
x
x
pOH = 14 − pH = 14 − 11.40 = 2.60
[OH ] = 10
−
Kb
− pOH
= 10 − 2.60 = 2.5 × 10 −3
[HOC6 H 5 ][OH − ] =
=
[OC H ]
−
6
7.
5
(
)
2
x2
2.5 × 10 −3
=
= 6.4 × 10 −5
−3
0.10 − x 0.10 − 2.5 × 10
K w 1.0 × 10 −14
Ka =
=
= 1.6 × 10 −10
−5
K b 6.4 × 10
What is the pH at the end of the following neutralization reactions?
a)
50.00 mL of 0.10 M CH3COOH combined with 50.00 mL of 0.10 M NaOH
see #5; pH = 8.7
b)
50.00 mL of 0.10 M NaOH combined with 50.00 mL of 0.1 M HCl
H+ + OH- Æ H2O
Since we begin with equal moles of both the strong acid and the
strong base, they will neutralize each other. The final pH = 7
c)
50.0 mL of 0.10 M NH3 combined with 50.00 mL of 0.10 M HCl
This is numerically exactly the same as part a. The strong acid will
be consumed entirely by reacting with NH3 to form the conjugate
acid (NH4+). The conjugate acid will then react with water to yield an
acidic solution.
So 0.0050 moles of NH4+ will form, with a concentration of 0.050M
and then react with water according to the equation NH4+ + H2O '
NH3 + H3O+
Initial
Change
Equil.
[NH4+]
0.050
-x
0.050 – x
K a = 5.6 × 10 −10 =
[NH3]
0
+x
x
[NH 3 ][H 3O + ] ≈
[NH ]
+
4
x ≈ 5.3 × 10 −6
pH = − log(5.3 × 10 −6 ) = 5.3
x2
0.050
[H3O+]
0
+x
x
8.
Four different bases, all at 0.10 M concentrations and 1.0 L volumes, are reacted
with 100 mL of 1.00 M HNO3. What are the pH values of the solutions after the
reaction?
0.10 mol CH 3 COOH
mol base = 1.0 L ×
= 0.10 mol
1L solution
1.00 mol H +
mol H + = 0.100 L ×
= 0.10 mol
1L solution
a)
CH3NH2 (Kb = 4.4 x 10-4)
CH3NH2 + HNO3 Æ CH3NH3+ + NO3This initial reaction produces 0.10 mol CH3NH3+ (weak acid) with a
concentration of 0.10 mol / 1.1 L = 0.0909 M
By now, we should see the general trend (without drawing out the
full ICE table), and solve immediately for the [H3O+]
K
1.0 × 10 −14
Ka = w =
= 2.3 × 10 −11
Kb
4.4 × 10 − 4
x2
0.0909
+
[ H 3O ] = x ≈ 1.4 × 10 −6
K a = 2.3 × 10 −11 ≈
(
)
pH = − log 1.4 × 10 −6 = 5.8
b)
NH3 (Kb = 1.8 x 10-5)
K w 1.0 × 10 −14
x2
−10
Ka =
=
= 5.6 × 10 ≈
K b 1.8 × 10 −5
0.0909
[ H 3O + ] = x ≈ 7.1 × 10 −6
(
)
pH = − log 7.1 × 10 −6 = 5.1
c)
C5H5N (Kb = 1.7 x 10-9)
K
1.0 × 10 −14
x2
−6
Ka = w =
=
5
.
9
×
10
≈
K b 1.7 × 10 −9
0.0909
[ H 3O + ] = x ≈ 7.3 × 10 − 4
(
)
pH = − log 7.3 × 10 − 4 = 3.1
d)
NaOH
Strong acid + Strong base Æ H2O (assuming equal moles of each)
pH = 7
Polyprotic acids are those with more than one acidic proton. One example is arsenic
acid, H3AsO4, a triprotic acid. It has three equilibrium expressions associated with its
reaction with water:
H3AsO4 + H2O ' H2AsO4- + H3O+
Ka1 = 5.0 x 10-3
H2AsO4- + H2O ' HAsO42- + H3O+
Ka2 = 8.0 x 10-8
HAsO42- + H2O ' AsO43- + H3O+
Ka3 = 6.0 x 10-10
It is a weak acid, whose ionization constants decrease markedly for the second and
third dissociation reactions. The concentrations of each of the four species can be
calculated as follows:
The value of Ka1 will be used to calculate the concentration of H3AsO4, H2AsO4- and
H3O+. These concentrations will not be affected by the other equilibria.
The value of Ka2 will be used to calculate the concentration of HAsO42-, using the
previously calculated value of H2AsO4- and H3O+.
The value of Ka3 will be used to calculate the concentration of AsO43- using the
previously calculated value of HAsO42- and the concentration of H3O+, from Ka1.
9.
If we have a 5.0 M solution of H3AsO4, what are the concentrations of all of the
species present?
a)
First, calculate the following concentrations using the first ionization:
[H2AsO4-] = _0.158_
[H3AsO4] = __5.0__
Initial
Change
Equil.
[H2AsO4-]
0
+x
x
[H3AsO4]
5.0
-x
5.0 – x
K a1 = 5.0 × 10 −3 =
[H
2
][
x ≈ 0.158
b)
]
AsO4− H 3O +
x2
≈
[H 3 AsO4 ]
5.0
What is the pH of this solution?
pH = − log(0.158) = 0.80
[H3O+] = _0.158_
[H3O+]
0
+x
x
c)
Calculate the [HAsO42-] by solving for Ka2.
Ka2 = 8.0 x 10-8 = [H3O+][HAsO42-]
[H2AsO4-]
Initial
Change
Equil.
[H2AsO4-]
0.158
-y
0.158 – y
K a 2 = 8.0 × 10 −8 =
Use [H2AsO4-] and [H3O+] from part a)
[HAsO4-2]
0
+y
y
[H3O+]
0.158
+y
0.158 + y
[HAsO ][H O ] = y(0.158 + y )
0.158 − y
[H AsO ]
−2
4
+
3
−
4
2
0.158 y
8.0 × 10 −8 ≈
0.158
−8
(y ≈ K a2 )
y ≈ 8.0 × 10
[HAsO42-] = _8.0x10-8____
d)
Calculate the [AsO43-] by solving for Ka3.
-10
Ka3 = 6.0 x 10
Initial
Change
Equil.
= [H3O ][AsO4 ]
[HAsO42-]
[HAsO4-2]
8.0x10-8
-z
8.0x10-8 – z
K a 3 = 6.0 × 10 −10 =
6.0 × 10 −10 ≈
Use [HAsO42-] from part c)
and [H3O+] from part a)
3-
+
[AsO4-3]
0
+z
z
[H3O+]
0.158
+z
0.158 + z
[AsO ][H O ] = z(0.158 + z )
[HAsO ] 8.0 × 10 − z
−3
4
+
3
−2
4
0.158 z
8.0 × 10 −8
z ≈ 3.0 × 10 −16
[AsO43-] = _3.0x10-16_____
−8
10.
Ascorbic acid, H2C6H6O6 is a diprotic acid, with Ka1 = 1.0 x 10-5 and
Ka2 = 5.0 x 10-12.
It is often abbreviated as H2Asc. Using this abbreviation:
a)
b)
b)
Write out the equilibria of this acid with water.
H2Asc + H2O ' HAsc- + H3O+
Ka1 = 1.0x10-5
HAsc- + H2O ' Asc-2 + H3O+
Ka2 = 5.0x10-12
Write all of the species that will exist in a 0.500 M solution of this weak
acid, and label them as acids, bases or both
species
acid/base
[species]e
i)
H2Asc
acid
ii)
HAsc-
both
iii)
Asc-2
base
iv)
H2O
both
v)
H3O+
acid (both?)
vi)
OH-
base (both?)
Calculate the equilibrium concentrations of all of these species (except H2O) and
enter them in the table above.
Using the previous results…
[H2Asc]eq ≈ [H2Asc]0 ≈ 0.500
[HAsc-] = [H3O+] ≈ x
K a1 = 1.0 × 10 −5 =
[HAsc ][H O ] ≈
−
+
3
[H 2 Asc]
x2
0.500
x ≈ 2.2 × 10 −3
[Asc-2] ≈ y ≈ Ka2
K a 2 = 5.0 × 10 −12
This leaves only [OH-], which can be found using the product of
hydronium and hydroxide ions: [H3O+][OH-] = 1.0x10-14
1.0 × 10 −14 1.0 × 10 −14
= 4.5 × 10 −12
OH − =
=
2.2 × 10 −3
H 3O +
[
] [
]