Sulphurous acid, H 2 SO 3 , is a weak acid capable of providing two H + ions. (a) What is the pH of a 0.45M solution of H 2 SO 3 ? (b) What is the equilibrium concentration of the sulphate ion SO 3 2− , in the 0.45M solution of H 2 SO 3 ?
Sulphurous acid, H 2 SO 3 , is a weak acid capable of providing two H + ions. (a) What is the pH of a 0.45M solution of H 2 SO 3 ? (b) What is the equilibrium concentration of the sulphate ion SO 3 2− , in the 0.45M solution of H 2 SO 3 ?
From the Ka1 and Ka2 values, Ka2 is smaller than the Ka1.
Therefore, H3O+ is almost produced entirely from Ka1.
Let’s calculate the H3O+ from Ka1.
First ionization:H2SO3(aq) + H2O(aq)⇌H3O+(aq) + HSO3-(aq)Equilibrium expression:Ka1 = [HSO3-][H3O+][H2SO3]Ka1= 1.2×10-2
Enter the ICE table the concentrations before equilibrium is established, the change that occurs as the reaction proceeds to equilibrium and the concentrations when equilibrium has been achieved.
H2SO3(aq) + H2O (aq)⇌ H3O+(aq) + HSO3-(aq)I 0.45M -- -- --C -x -- +x +xE (0.45-x) -- x x
The equilibrium concentration has to be determined for the sulphate ion SO32- in the 0.45M solution of H2SO3.
Concept Introduction:
Equilibrium constants:
The equilibrium constant is used to quantitative measurement of the strength of the acid and bases in the water.
Ka is acid constant for equilibrium reactions.
HA + H2O⇌H3O++ A-Ka= [H3O+][A-][HA]
Kb is base constant for equilibrium reaction.
BOH + H2O⇌B++ OH-Ka= [B+][OH-][BOH]
If a molecule donate only one hydrogen atom then the acid is called a monoprotic acid.
Example: HCl, HBr and HClO3
If each molecule can donate two or more hydrogen ions, the acid is a polyprotic acid.
A diprotic acid, such as sulfuric acid, H2SO4 has two acidic hydrogen atoms.
Some acids, such as phosphoric acid, H3PO4 are triprotic acids.
Polyprotic bases:
Can accept more than on proton (H+). So, it can react with water many times to produce many hydroxide ions (OH−).
Diprotic bases:
Can accept two protons.
Example: Sulphate ion SO4(aq)2-
Answer to Problem 73PS
The equilibrium concentration of the sulphate ion SO32- in the 0.45M solution of H2SO3 is 6.28×10-8M
Explanation of Solution
Second ionization:HSO3-(aq) + H2O(aq)⇌H3O+(aq) + SO32-(aq)Equilibrium expression:Ka2= [H3O+][SO32-][HSO3-]Ka2= 6.2×10-8
Enter the ICE table the concentrations before equilibrium is established, the change that occurs as the reaction proceeds to equilibrium and the concentrations when equilibrium has been achieved.
6.2×10-8 = y (6.77×10-2 + y)(6.77×10-2- y)Y is so small compared to 6.77×10-2So, all values are almost equal to 6.77×10-26.2×10-8 = y(6.77×10-2)(6.77×10-2) y = 6.2×10-8
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