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Recently Added Formulas in Chemistry
·
Hydrolysis
·
Ionization
·
pH scale
·
Acid-base Reaction
·
Strength of Acid and bases
Additional Formulas
·
Acid and Base
·
Bronsted/Lowry theory
·
Strength of Acid and bases
·
Acid-base Reaction
·
pH scale
·
Buffers
·
Ionization
·
Hydrolysis
Current Location
>
Formulas in Chemistry
>
Acid and Bases
> Hydrolysis
Hydrolysis
Definition:
When a salt of acid or base is dissolved in water, interaction occurs between ions of a salt and water is called hydrolysis.
Salts are divided into four types.
Salts obtained from strong acid and strong base like KCl, Na
2
SO
4
, and KNO
3
.
After hydrolysis the solution remains neutral.
Salts obtained from weak acid and strong base like CH
3
COONa, Na
2
CO
3
, and K
3
PO
4
.
After hydrolysis the solution remains alkaline.
Salts obtained from strong acid and weak base like NH
4
Cl, CuSO
4
, and Al (NO
3
)
3
.
After hydrolysis the solution remains acidic.
Salts obtained from weak acid and weak base like aluminium acetate and ammonium formate.
Hydrolysis of salt obtained from Weak acid and Strong base:
Let we have MA a salt of weak acid and strong base. When MA is dissolved in water, the interaction occurs between A
-
ion and water molecules.
A
-
+ H
2
O ↔ HA + OH
-
At equilibrium Keq =
[HA][OH
-
]
[A
-
][H
2
O]
Keq [H
2
O] = Kh =
[HA][OH
-
]
[A
-
] (Kh = hydrolysis constant)………………… (1)
We have ionic product of water Kw = [H
+
] [OH
-
]
And ionization constant of weak acid Ka =
[H
+
] [A
-
]
[HA]
Kw/Ka = [H
+
] [OH
-
]/
[H
+
] [A
-
]
[HA]
=
[HA] [OH
-
]
[A
-
]
= Kh …………………………………………………………..(from equation 1)
Kh = Kw/Ka
Hydrolysis constant is related to ionic product of water and ionization constant of the acid.
Thus Kh =
Kw
=
[HA] [OH
-
]
=
[OH
-
]
2
Ka [A
-
] C
As the amount of [HA] and [OH
-
] formed by hydrolysis is equal and [A
-
] = original concentration of salt in moles/lit
Kh =
Kw
=
[OH
-
]
2
Ka C
Hydrolysis of salt obtained from Strong acid and Weak base:
Let we have MA a salt of strong acid and weak base. When MA is dissolved in water, the interaction occurs between M
+
ion and water molecules.
M
+
+ H
2
O ↔ H
+
+ MOH
At equilibrium Keq =
[H
+
][MOH]
[M
+
][H
2
O]
Keq [H
2
O] = Kh =
[H
+
][MOH]
[M
+
] (Kh = hydrolysis constant)………………… (2)
We have ionic product of water Kw = [H
+
] [OH
-
]
And ionization constant of weak base Kb =
[M
+
] [OH
-
]
[MOH]
Kw/Kb = [H
+
] [OH
-
]/
[M
+
] [OH
-
]
[MOH]
=
[H
+
] [MOH]
[M
+
]
= Kh (from equation 2)
Kh = Kw/Kb
Hydrolysis constant is related to ionic product of water and ionization constant of the acid.
Thus Kh =
Kw
=
[H
+
] [MOH]
=
[H
+
]
2
Kb [M
+
] C
As the amount of [H
+
] and [MOH] formed by hydrolysis is equal and [M
+
] = original concentration of salt in moles/lit
Kh =
Kw
=
[H
+
]
2
Ka C
Hydrolysis of salt obtained from Weak acid and Weak base:
Let we have MA a salt of weak acid and weak base. When MA is dissolved in water, the interaction occurs between M
+
, A
-
ion and water molecules.
M
+
+ A
-
+ H
2
O ↔ HA + MOH
At equilibrium Keq =
[HA][MOH]
[M
+
][A
-
][H
2
O]
Keq [H
2
O] = Kh =
[HA][MOH]
[M
+
][A
-
] (Kh = hydrolysis constant)…………………(3)
We have ionic product of water Kw = [H
+
] [OH
-
]
And ionization constant of weak acid Ka =
[H
+
] [A
-
]
[HA]
And ionization constant of weak base Kb =
[M
+
] [OH
-
]
[MOH]
Kw
=
[H
+
] [OH
-
] [HA] [MOH]
Ka x Kb [H
+
] [A
-
] [M
+
] [OH
-
]
=
[HA] [MOH]
[A
-
] [M
+
]
= Kh (from equation 3)
Kh = __
Kw__
Ka x Kb
Calculation:
Exp 1:
Calculate the hydrolysis constant of acetic acid when its ionization constant is 1.75 x 10
-5
Reason:
Ka = 1.75 x 10
-5
Kw = 1.0 x 10
-14
Kh = ?
Kh =
Kw
Ka
=
1.0 x 10
-14
1.75 x 10
-5
= 5.7 x 10
-10
Exp 2:
Calculate the pH of an aqueous solution of 0.01M acetic acid when its ionization constant is 1.75 x 10
-5
Reason:
Ka = 1.75 x 10
-5
Kw = 1.0 x 10
-14
C = 0.01M
pH = ?
Kh =
Kw
=
[OH
-
]
2
Ka C
1.0 x 10
-14
=
[OH
-
]
2
1.75 x 10
-5
0.01
[OH
-
]
2
= 5.7 x 10
-12
[OH
-
] = 2.39
pOH = - log [OH
-
]
= - log [2.39]
= - (-5.62)
= 5.62
pH = 14- pOH
= 14 – 5.62
= 8.38
Exp 3:
Calculate the pH of 0.02M solution of methyl amine when its ionization constant is 5.00 x 10
-4
Reason:
Kb = 5.00 x 10
-4
Kw = 1.0 x 10
-14
C = 0.02M
pH = ?
Kh =
Kw
=
[H
+
]
2
Kb C
1.0 x 10
-14
=
[H
+
]
2
5.0 x 10
-4
0.02
[H
+
]
2
= 0.4. x 10
-12
[H
+
] = 6.32 x 10
-7
pH = - log [H
+
]
= - log [6.32 x 10
-7
]
= 6.19
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