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Worksheet 20 – Polyprotic Acids and Salt Solutions
K a Acid Base K b strong acid HNO 3 , HI, HCl, etc NO 3 - , I - , Cl-, etc negligible basicity 1.3 x 10 -2^ HSO 4 -^ SO 4 2-^ 7.7 x 10 - 7.1 x 10 -4^ HNO 2 NO 2 -^ 1.4 x 10 - 6.8 x 10 -4^ HF F-^ 1.5 x 10 - 1.8 x 10 -5^ CH 3 COOH CH 3 COO -^ 5.6 x 10 - 4.5 x 10 -7^ H 2 CO 3 HCO 3 -^ 2.3 x 10 - 9 x 10-8^ H 2 S HS-^ 1.1 x 10 - 5.6 x 10 -10^ NH 4 +^ NH 3 1.8 x 10 - 6.2 x 10 -10^ HCN CN-^ 1.6 x 10 - 4.7 x 10 -11^ HCO 3 -^ CO 3 2-^ 1.8 x 10 - 1 x 10-17^ HS-^ S2-^ 1 x 10^3 negligible acidity Li 2 O, NaOH O2-, OH-^ 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.
- 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: KCH 3 COO was formed in the reaction between KOH and CH 3 COOH
KOH is a strong base. Its conjugate acid, K+^ , has negligible acidity and will leave the pH at 7.00, a neutral solution.
CH 3 COOH is a weak acid , making its conjugate base, CH 3 COO -^ a relatively strong base. It will produce a basic solution.
a) KF b) KCN
basic basic
c) NaNO 3 d) RbI
neutral neutral
e) NH 4 NO 3 f) Na 2 CO (^3)
acidic basic
- Rank the following 0.1 M aqueous salt solutions in order of increasing pH. (Hint: write out the reactions of the salts + water)
a) KNO 3 K 2 SO 4 K 2 S neutral basic (Kb = 7.7x10-13^ ) basic (Kb = 1x10^3 ) strongest base
pH rank: KNO 3 < K 2 SO 4 < K 2 S
b) NH 4 NO 3 NaHSO 4 NaHCO 3 Na 2 CO (^3) acidic acidic basic basic Ka = 5.6x10-10^ Ka = 1.3x10-2^ Kb = 2.3x10-8^ Kb = 1.8x10- strongest acid strongest base
pH rank: NaHSO 4 < NH 4 NO 3 < NaHCO 3 < Na 2 CO 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
-
< Z
-
The conjugate acids will have the reverse trend: HZ < HY < HX
- 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 -
CN-^ + H 2 O ' HCN + OH-
[CN-] [HCN] [OH-]
Initial 0.10 0 0 Change -x +x +x Equil. 0.10 – x x x
[ ][ ] [ ] 3
2 5
−
−
− −
≈ ×
= × = ≈
x
x CN
HCN OH
K (^) b
log( 1. 3 103 ) 2. 9 = − =
= − × − =
pH
pOH
- A 0.10 M solution of KOC 6 H 5 has a pH of 11.40. Calculate the K a value for HOC 6 H 5. [OC 6 H 5 - ] [HOC 6 H 5 ] [OH-] Initial 0.10 0 0 Change -x +x +x Equil. 0.10 – x x x
[ ] 10 102.^602. 5 103
− = − = − = × −
OH^ pOH
pOH pH
[ ][ ]
[ ]
10 5
14
5 3
2 32
6 5
6 5
− −
−
− −
− −
−
= ×
×
×
= ×
− ×
×
b
w a
b
K
K
K
x
x OCH
HOC H OH
K
- What is the pH at the end of the following neutralization reactions?
a) 50.00 mL of 0.10 M CH 3 COOH combined with 50.00 mL of 0.10 M NaOH see #5; pH = 8.
b) 50.00 mL of 0.10 M NaOH combined with 50.00 mL of 0.1 M HCl H+^ + OH-^ Æ H 2 O
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 NH 3 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 NH 3 to form the conjugate acid (NH 4 +^ ). The conjugate acid will then react with water to yield an acidic solution.
So 0.0050 moles of NH 4 +^ will form, with a concentration of 0.050M and then react with water according to the equation NH 4 +^ + H 2 O ' NH 3 + H 3 O +
[NH 4 +^ ] [NH 3 ] [H 3 O +^ ] Initial 0.050 0 0 Change -x +x +x Equil. 0.050 – x x x
[ ][ ]
[ ]
6
2
4
(^1033)
−
−
≈ ×
= × = ≈
x
x NH
NH HO
K (^) a
pH = −log( 5. 3 × 10 −^6 )= 5. 3
Four different bases, all at 0.10 M concentrations and 1.0 L volumes, are reacted with 100 mL of 1.00 M HNO 3. What are the pH values of the solutions after the reaction?
10 mol 1 Lsolution
00 molH molH 0. 100
10 mol 1 Lsolution
10 molCHCOOH mol base 1. 0 3
= × =
= × =
L
a) CH 3 NH 2 ( K b = 4.4 x 10-4^ ) CH 3 NH 2 + HNO 3 Æ CH 3 NH 3 +^ + NO 3 - This initial reaction produces 0.10 mol CH 3 NH 3 +^ (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 [H 3 O +^ ]
log ( 1. 4 10 ) 5. 8
[ ] 1. 4 10
6
6 3
2 11
11 4
14
=− × =
= ≈ ×
= × ≈
= ×
×
×
−
−
− −
−
pH
HO x
x K
K
K
K
a
b
w a
b) NH 3 ( K b = 1.8 x 10-5^ )
log ( 7. 1 10 ) 5. 1
[ ] 7. 1 10
6
6 3
2 10 5
14
=− × =
= ≈ ×
= × ≈
×
×
−
− −
−
pH
HO x
x K
K
K
b
w a
c) C 5 H 5 N ( K b = 1.7 x 10-9^ )
log ( 7. 3 10 ) 3. 1
[ ] 7. 3 10
4
4 3
2 6 9
14
=− × =
= ≈ ×
= × ≈
×
×
−
− −
−
pH
HO x
x K
K
K
b
w a
d) NaOH Strong acid + Strong base Æ H 2 O (assuming equal moles of each) pH = 7
c) Calculate the [HAsO 4 2-] by solving for K a.
K a2 = 8.0 x 10-8^ = [H 3 O +^ ][HAsO 4 2-] [H 2 AsO 4 - ]
[H 2 AsO 4 - ] [HAsO 4 -2^ ] [H 3 O +^ ] Initial 0.158 0 0. Change -y +y +y Equil. 0.158 – y y 0.158 + y
[ ][ ] [ ]
( )
(^8) ( 2 )
8
2 4
3
2 (^84) 2
a
a
y y K
y
y
y y H AsO
HAsO HO K
≈ × ≈
× ≈
= × = =
−
−
−
− + −
[HAsO 4 2-] = _ 8.0x10 -8 ____
d) Calculate the [AsO 4 3-] by solving for K a.
K a3 = 6.0 x 10-10^ = [H 3 O +^ ][AsO 4 3-] [HAsO 4 2-]
[HAsO 4 -2^ ] [AsO 4 -3^ ] [H 3 O +^ ] Initial 8.0x10 -8^0 0. Change -z +z +z Equil. 8.0x10 -8^ – z z 0.158 + z
[ ][ ] [ ]
( )
16
8
10
2 8 4
3
3 10 4 3
−
−
−
− −
− + −
≈ ×
×
× ≈
× −
= × = =
z
z
z
z z HAsO
AsO HO K (^) a
[AsO 4 3-] = _ 3.0x10 -16 _____
Use [H 2 AsO 4 - ] and [H 3 O +^ ] from part a)
Use [HAsO 4 2-] from part c) and [H 3 O +^ ] from part a)
- Ascorbic acid, H 2 C 6 H 6 O 6 is a diprotic acid, with K a1 = 1.0 x 10-5^ and K a2 = 5.0 x 10-^.
It is often abbreviated as H 2 Asc. Using this abbreviation:
a) Write out the equilibria of this acid with water.
H 2 Asc + H 2 O ' HAsc-^ + H 3 O +^ Ka1 = 1.0x10-
HAsc -^ + H 2 O ' Asc-2^ + H 3 O +^ Ka2 = 5.0x10-
b) 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) H 2 Asc acid
ii) HAsc -^ both
iii) Asc-2^ base
iv) H 2 O both
v) H 3 O +^ acid (both?)
vi) OH-^ base (both?)
b) Calculate the equilibrium concentrations of all of these species (except H 2 O) and enter them in the table above. Using the previous results… [H 2 Asc] (^) eq ≈ [H 2 Asc] 0 ≈ 0.
[HAsc-] = [H 3 O +^ ] ≈ x [ ][ ] [ ] 3
2
2
(^53) 1
- 2 10
−
− + −
≈ ×
= × = ≈
x
x H Asc
HAsc HO K (^) a
[Asc -2^ ] ≈ y ≈ Ka 12 2 5.^010
= × −
Ka
This leaves only [OH-], which can be found using the product of hydronium and hydroxide ions: [H 3 O +^ ][OH-] = 1.0x10-
[ ] [ ]
12 3
14
3
14
- 5 10
- 2 10
−
−
− − (^) = × ×
×
×
HO
OH
