**Answer the following :**

Derive Ostwald's dilution law for the CH_{3}COOH.

#### Solution

**i. **

Consider an equilibrium of weak acid CH_{3}COOH that exists in solution partly as the undissociated species CH_{3}COOH and partly H^{+} and CH_{3}COO^{–} 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 CH_{3}COOH is initially present in volume V dm^{3} 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 [CH_{3}COOH] = `(1 - α)/"V"` mol dm^{-3},

[H^{+}] = [CH_{3}COO^{-}] = `α/"V"` mol dm^{-3}

**v. **

Substituting these in equation (1),

K_{a }= `(α / "V" α / "V") /((1 - α) / "V") = α^2/(1 - α "V")` ...(2)

**vi. **

If c is the initial concentration of CH_{3}COOH in mol dm^{–3} and V is the volume in dm^{3} mol^{–1} then c = 1/V. Replacing 1/V in equation (2) by c,

we get

K_{a }= `(α^2"c")/(1 - α)` ...(3)

**vii.**

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

K_{a }= `α^2/"V"` and K_{a }= α^{2}c ...(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 (CH_{3}COOH) 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.