# Answer the following : Derive Ostwald's dilution law for the CH3COOH. - Chemistry

Derive Ostwald's dilution law for the CH3COOH.

#### Solution

i.

Consider an equilibrium of weak acid CH3COOH that exists in solution partly as the undissociated species CH3COOH and partly H+ and CH3COO ions. Then

$\ce{CH3COOH_{(aq)} ⇌ H^+_{ (aq)} + CH3COO^-_{ (aq)}}$

ii.

The acid dissociation constant is given as:

"K"_"a" = (["H"^+]["CH"_3"COO"^-])/[["CH"_3"COOH"]]  ....(1)

iii.

Suppose 1 mol of acid CH3COOH is initially present in volume V dm3 of the solution. At equilibrium, the fraction dissociated would be α, where α is the degree of dissociation of the acid. The fraction of an acid that remains undissociated would be (1 - α).

 $\ce{CH3COOH_{(aq)}⇌ H^+_{ (aq)} + CH3COO^-_{ (aq)}}$ The amount present at equilibrium (mol) (1 - α) α α Concentration at equilibrium (mol dm−3) (1 - α)/"V" α/"V" α/"V"

iv.

Thus, at equilibrium [CH3COOH] = (1 - α)/"V" mol dm-3,

[H+] = [CH3COO-] = α/"V" mol dm-3

v.

Substituting these in equation (1),

K= (α / "V" α / "V") /((1 - α) / "V") = α^2/(1 - α  "V")  ...(2)

vi.

If c is the initial concentration of CH3COOH in mol dm–3 and V is the volume in dm3 mol–1 then c = 1/V. Replacing 1/V in equation (2) by c,

we get

K= (α^2"c")/(1 - α)  ...(3)

vii.

For the weak acid CH3COOH, α is very small, or (1 − α) ≅ 1. With this equation (2) and (3) becomes:

K= α^2/"V" and K= α2c  ...(4)

α = sqrt("K"_"a")/"c" or α = sqrt("K"_"a". "V")  ...(5)

Equation (5) implies that the degree of dissociation of a weak acid (CH3COOH) is inversely proportional to the square root of its concentration or directly proportional to the square root of the volume of the solution containing 1 mol of the weak acid.

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#### APPEARS IN

Balbharati Chemistry 12th Standard HSC Maharashtra State Board
Chapter 3 Ionic Equilibria
Exercise | Q 4.01 | Page 62