Derive ostwald's dilute law for weak acid ch3cooh
Answers
Answered by
48
When acetic acid is dissolved in water, it dissociates partially into H+ or CH3COO– and ions and the following equilibrium is obtained.
CH3COOH + H2O<-----> CH3COO– + H3O+
Applying law of chemical equilibrium, In dilute solution, [H2O] is constant. The product of K and constant [H2O] is denoted as Ka, the ionization constant or dissociation constant of the acid is,Ka = [CH3COO–]×[H3O+]/[CH3COOH] …..(a)
The fraction of total number of molecules of an electrolyte which ionise into ions is known as degree of dissociation/ionisation α.Let 'C" moles L–1 be the initial molar concentration of acetic acid and α be the degree of dissociation. Therefore, molar concentrations of different species before the ionisation and at equilibrium point are expressed as follows:
CH3COOH (aq)<----> H3O+ (aq) + CH3COO-(aq)
Initial conc. C 0 0
Conc. at equil C(1- α) Cα Cα
Substituting the values of the equilibrium concentrations in equation (a),we get
Ka = Cα.Cα/C(1–α)
= C2α2/C(1–α)
= Cα2/1–α
Since for weak electrolytes, the value of α is very small and can be neglected in comparison to 1 i.e., .1 – α = 1.Hence, we get
Kα = Cα2 or α = √Ka/C
Furher, if V is the volume of the solution in litres containing 1 mole of the electrolyte,
C = 1/V.
Hence we have
α =√KaV ......(b)
Similarly, for a weak base like NH4OH, we haveα = √Kb/C = √KbV ........(c)
From equations (b) and (c),we can conclude the following statement.
In case of a weak electrolyte at a given temperature, the degree of ionization is inversely proportional to the square root of the molar concentration or directly proportional to square root of the volume of the solution which contains one mole of the electrolyte. This is called Ostwald’s dilution Law.
CH3COOH + H2O<-----> CH3COO– + H3O+
Applying law of chemical equilibrium, In dilute solution, [H2O] is constant. The product of K and constant [H2O] is denoted as Ka, the ionization constant or dissociation constant of the acid is,Ka = [CH3COO–]×[H3O+]/[CH3COOH] …..(a)
The fraction of total number of molecules of an electrolyte which ionise into ions is known as degree of dissociation/ionisation α.Let 'C" moles L–1 be the initial molar concentration of acetic acid and α be the degree of dissociation. Therefore, molar concentrations of different species before the ionisation and at equilibrium point are expressed as follows:
CH3COOH (aq)<----> H3O+ (aq) + CH3COO-(aq)
Initial conc. C 0 0
Conc. at equil C(1- α) Cα Cα
Substituting the values of the equilibrium concentrations in equation (a),we get
Ka = Cα.Cα/C(1–α)
= C2α2/C(1–α)
= Cα2/1–α
Since for weak electrolytes, the value of α is very small and can be neglected in comparison to 1 i.e., .1 – α = 1.Hence, we get
Kα = Cα2 or α = √Ka/C
Furher, if V is the volume of the solution in litres containing 1 mole of the electrolyte,
C = 1/V.
Hence we have
α =√KaV ......(b)
Similarly, for a weak base like NH4OH, we haveα = √Kb/C = √KbV ........(c)
From equations (b) and (c),we can conclude the following statement.
In case of a weak electrolyte at a given temperature, the degree of ionization is inversely proportional to the square root of the molar concentration or directly proportional to square root of the volume of the solution which contains one mole of the electrolyte. This is called Ostwald’s dilution Law.
Answered by
20
Answer:
In case of a weak electrolyte at a given temperature, the degree of ionization is inversely proportional to the square root of the molar concentration or directly proportional to square root of the volume of the solution which contains one mole of the electrolyte. This is called Ostwald's dilution Law.
Attachments:
Similar questions