Polyprotic Acids and Salts in Chemistry
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Polyprotic Acids
Homework
Complete the AP practice test at the end of
Chapter 14 Acid and Bases
It is in between page 598-599
Polyprotic Acids
A polyprotic acid has more than one
ionizable proton (H
+
).
e.g. H
2
SO
4
, H
2
CO
3
, H
3
AsO
4
Each successive K
a
value get smaller (K
a1
> K
a2
> K
a3
). Therefore, the first
dissociation step makes the most
significant contribution to the
equilibrium concentration of [H
+
].
Sulfuric acid is unique in being a
strong acid in its first dissociation
step and a weak acid in its second
step.
LeCh
âtelier
For 1.0 M or larger solutions of sulfuric
acid, the large concentration of H
3
O
+
from the first dissociation step
represses the second step, which can be
neglected as a contributor of H
3
O
+
ions.
For dilute solutions of sulfuric acid, the
second step does make a significant
contribution, ICE tables must be used to
obtain the
total H
3
O
+
concentration.
Problem
Calculate the pH of .50 M H
2
SO
4
H
2
SO
4
HSO
4
-
+ H
+
K
a
is very
large
HSO
4
-
⇌
SO
4
2-
+ H
+
K
a
= 1.2x10
-2
The main difference from sulfuric acid from
other problems will be the first step.
0.50 M H
2
SO
4
means .50 M H
+
and .5 M
HSO
4
-
is formed.
Now to the second dissociation
Cont.
HSO
HSO
4
4
-
-
SO
SO
4
4
2-
2-
+ H
+ H
+
+
I .50 M .50 M
I .50 M .50 M
C -x +x +x
C -x +x +x
E .50 –x x .50 + x
E .50 –x x .50 + x
K
K
a
a
= x(.5+x) / (.5-x) = 1.2 x10
= x(.5+x) / (.5-x) = 1.2 x10
-2
-2
x = .011462
x = .011462
[HSO
[HSO
4
4
-
-
]= .49 M [SO
]= .49 M [SO
4
4
2-
2-
]= .011 M
]= .011 M
[H
[H
+
+
] = .51 M
] = .51 M
pH = .29
pH = .29
Problem
Calculate the pH, [H
Calculate the pH, [H
+
+
], [HSO
], [HSO
4
4
-
-
], [SO
], [SO
4
4
2-
2-
] of
] of
.20 M H
.20 M H
2
2
SO
SO
4
4
Calculate the pH of a 5.0 M H
Calculate the pH of a 5.0 M H
3
3
PO
PO
4
4
solution
solution
and the equilibrium concentrations of the
and the equilibrium concentrations of the
species H
species H
3
3
PO
PO
4
4
, H
, H
2
2
PO
PO
4
4
-
-
, HPO
, HPO
4
4
2-
2-
, and PO
, and PO
4
4
3-
3-
.
.
K
K
a1
a1
= 7.5 x 10
= 7.5 x 10
-3
-3
K
K
a2
a2
= 6.2 x 10
= 6.2 x 10
-8
-8
K
K
a3
a3
= 4.8 x 10
= 4.8 x 10
-13
-13
Salts
Salts and pH
Salts are made of conjugate acids and bases.
Salts are made of conjugate acids and bases.
The conjugate acid/bases of strong acids or
The conjugate acid/bases of strong acids or
base will be neutral, excluding H
base will be neutral, excluding H
2
2
SO
SO
4
4
.
.
To be a strong acid or base, the conjugate
To be a strong acid or base, the conjugate
base or acid must have no affinity for
base or acid must have no affinity for
protons or hydroxide.
protons or hydroxide.
That is to say it won’t run the reverse
That is to say it won’t run the reverse
reaction under
reaction under
any
any
conditions.
conditions.
Salts of weak acids and bases
Conjugate bases/acids of weak acids and
Conjugate bases/acids of weak acids and
bases will run the reverse reactions.
bases will run the reverse reactions.
Conjugate base of weak acids are weak
Conjugate base of weak acids are weak
bases
bases
HF
HF
⇌
⇌
H
H
+
+
+ F
+ F
-
-
F
F
-
-
+ H
+ H
2
2
O
O
⇌
⇌
HF
HF
+ OH
+ OH
-
-
Conjugate acids of weak bases are weak
Conjugate acids of weak bases are weak
acids
acids
NH
NH
3
3
⇌
⇌
NH
NH
4
4
+
+
+ OH
+ OH
-
-
NH
NH
4
4
+
+
⇌
⇌
NH
NH
3
3
+ H
+ H
+
+
pH of salt
Salts are made by joining the conjugate
Salts are made by joining the conjugate
acids and bases
acids and bases
Would potassium chlorite be acidic or
Would potassium chlorite be acidic or
basic?
basic?
KClO
KClO
2
2
⇌
⇌
K
K
+
+
+ ClO
+ ClO
2
2
-
-
Potassium is neutral (KOH = strong base),
Potassium is neutral (KOH = strong base),
chlorite is
chlorite is
basic (HClO
basic (HClO
2
2
= weak acid)
= weak acid)
Therefore it is basic
Therefore it is basic
Question
Are these salts acidic or basic?
NH
4
Cl
NaCH
3
COO
Ca(NO
3
)
2
What about salts of a weak acid and
a weak bases…
If you have NH
4
F…
You have to look at the K
a
value of NH
4
+
and compare it to the K
b
value of F
-
to see
which weak acid/base is stronger.
The larger the K value the stronger the
acid/base.
14
K
a
and K
b
The K
The K
a
a
and K
and K
b
b
of a weak acid and its
of a weak acid and its
conjugate base or a weak base and its
conjugate base or a weak base and its
conjugate acid are related and easily
conjugate acid are related and easily
calculated from one another
calculated from one another
For a weak acid and its conjugate base or
For a weak acid and its conjugate base or
a weak base and its conjugate acid
a weak base and its conjugate acid
only
only
K
K
a
a
x K
x K
b
b
= K
= K
w
w
Problem
Calculate the K
Calculate the K
a
a
of NH
of NH
4
4
Cl solution. The K
Cl solution. The K
b
b
value for NH
value for NH
3
3
is 1.8 x 10
is 1.8 x 10
-5
-5
.
.
Calculate the K
Calculate the K
b
b
of NaF solution. The K
of NaF solution. The K
a
a
value for HF is 7.2 x 10
value for HF is 7.2 x 10
-4
-4
.
.
Hydrated metal ions
Charged metals ions also produce an acidic
solution.
The metal itself does not act as a Brønsted-
Lowry acid, but instead forms a hydrate that
acts as a Brønsted-Lowry acid.
Typically the higher the charge on the metal
ion, stronger the acidity of the hydrated ion.
Cu
2+
+ 5 H
2
O
2
O)
5
2+
Cu(H
2
O)
5
2+
⇌
CuOH(H
2
O)
4
2+
+ H
+
Acidity of hydrated metal ions
M = metal
For M
n+
, a small, highly charged ion,
M(H
2
O)
x
n+
+ H
2
O
⇌
⇌
M(H
2
O)
x-1
OH
(n-1)
+ H
3
O
+
Calculate the pH of a 0.010 M AlCl
3
solution.
The K
a
value for Al(H
2
O)
6
3+
is 1.4 x 10
-5
.
Exploring the concept of polyprotic acids, particularly focusing on sulfuric acid as a unique example with its dual dissociation steps. The discussion delves into the equilibrium concentrations of ions in solution, calculating pH values for different acid concentrations, and understanding the significance of the first dissociation step in polyprotic acids. Additionally, we examine the role of salts in maintaining pH balance, highlighting the characteristics of conjugate acids and bases. Through examples and explanations, this content provides a comprehensive overview of these key topics in chemistry.
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Presentation Transcript
Homework Complete the AP practice test at the end of Chapter 14 Acid and Bases It is in between page 598-599
Polyprotic Acids A polyprotic acid has more than one ionizable proton (H+). e.g. H2SO4, H2CO3, H3AsO4 Each successive Kavalue get smaller (Ka1 > Ka2> Ka3). Therefore, the first dissociation step makes the most significant contribution to the equilibrium concentration of [H+].
Sulfuric acid is unique in being a strong acid in its first dissociation step and a weak acid in its second step.
LeChtelier For 1.0 M or larger solutions of sulfuric acid, the large concentration of H3O+ from the first dissociation step represses the second step, which can be neglected as a contributor of H3O+ions. For dilute solutions of sulfuric acid, the second step does make a significant contribution, ICE tables must be used to obtain the total H3O+concentration.
Problem Calculate the pH of .50 M H2SO4 H2SO4 HSO4-+ H+ HSO4- SO42-+ H+ The main difference from sulfuric acid from other problems will be the first step. 0.50 M H2SO4means .50 M H+and .5 M HSO4-is formed. Now to the second dissociation Kais very large Ka= 1.2x10-2
Cont. I .50 M .50 M C -x +x +x E .50 x x Ka= x(.5+x) / (.5-x) = 1.2 x10-2 x = .011462 [HSO4-]= .49 M [SO42-]= .011 M [H+] = .51 M pH = .29 HSO4- SO42-+ H+ .50 + x
Problem Calculate the pH, [H+], [HSO4-], [SO42-] of .20 M H2SO4 Calculate the pH of a 5.0 M H3PO4solution and the equilibrium concentrations of the species H3PO4, H2PO4-, HPO42-, and PO43-. Ka1= 7.5 x 10-3 Ka2= 6.2 x 10-8 Ka3= 4.8 x 10-13
Salts and pH Salts are made of conjugate acids and bases. The conjugate acid/bases of strong acids or base will be neutral, excluding H2SO4. To be a strong acid or base, the conjugate base or acid must have no affinity for protons or hydroxide. That is to say it won t run the reverse reaction under any conditions.
Salts of weak acids and bases Conjugate bases/acids of weak acids and bases will run the reverse reactions. Conjugate base of weak acids are weak bases HF H++ F- F-+ H2O HF + OH- Conjugate acids of weak bases are weak acids NH3 NH4++ OH- NH4+ NH3+ H+
pH of salt Salts are made by joining the conjugate acids and bases Would potassium chlorite be acidic or basic? KClO2 K++ ClO2- Potassium is neutral (KOH = strong base), chlorite is basic (HClO2= weak acid) Therefore it is basic
Question Are these salts acidic or basic? NH4Cl NaCH3COO Ca(NO3)2
What about salts of a weak acid and a weak bases If you have NH4F You have to look at the Kavalue of NH4+ and compare it to the Kbvalue of F-to see which weak acid/base is stronger. The larger the K value the stronger the acid/base. 14
Ka and Kb The Ka and Kb of a weak acid and its conjugate base or a weak base and its conjugate acid are related and easily calculated from one another For a weak acid and its conjugate base or a weak base and its conjugate acid only Ka x Kb = Kw
Problem Calculate the Ka of NH4Cl solution. The Kb value for NH3 is 1.8 x 10-5. Calculate the Kb of NaF solution. The Ka value for HF is 7.2 x 10-4.
Hydrated metal ions Charged metals ions also produce an acidic solution. The metal itself does not act as a Br nsted- Lowry acid, but instead forms a hydrate that acts as a Br nsted-Lowry acid. Typically the higher the charge on the metal ion, stronger the acidity of the hydrated ion. Cu2+ + 5 H2O Cu(H2O)52+ Cu(H2O)52+ CuOH(H2O)42+ + H+
Acidity of hydrated metal ions M = metal For Mn+, a small, highly charged ion, M(H2O)xn++ H2O M(H2O)x-1OH(n-1) + H3O+ Calculate the pH of a 0.010 M AlCl3 solution. The Ka value for Al(H2O)63+ is 1.4 x 10-5.
