# A weak acid,HA has the concentration of C mol/dm and degree of dissociation ,β.Assuming Ka<<1, prove that β is inversely proportional to square root of C?

##### 1 Answer

#### Answer:

Here's how you can prove this.

#### Explanation:

Before doing anything else, make sure that you understand what **degree of dissociation** means.

As you know, *weak acids* and *weak bases* **do not** dissociate completely in aqueous solution to form ions.

This means that at a given temperature, the number of molecules of weak acid that will dissociate to produce hydronium ions, *dissociation constant*,

Simply put, **not all molecules** of acid will dissociate; some will do, but **most** will not. This is what *degree of dissociation* tells you, i.e. what *Fraction* of the acid molecules dissociate.

In your case, the degree of dissociation for the weak acid **for every** mole of **each** ion will be formed in solution.

For simplicity, I'll use hydrogen ions,

The dissociation of a weak acid looks like this

#"HA"_text((aq]) rightleftharpoons "H"_text((aq])^(+) + "A"_text((aq])^(-)#

As you know, the acid dissociation constant,

#K_a = ( ["H"^(+)] * ["A"^(-)])/(["HA"])#

You know that

#["HA"] = c#

Since we've established that the degree of dissociation tells you what fraction of the acid molecules dissociate, you can say that

#["H"^(+)] = beta * c" "# and#" "["A"^(-)] = beta * c#

Every **that dissociates** will produce

Plug this into the equation for

#K_a = ( beta * color(red)(cancel(color(black)(c))) * beta* c)/color(red)(cancel(color(black)(c))) = beta^2 * c#

Your goal now is to isolate

#beta^2 = K_a * 1/c#

Take the square root of both sides to get

#beta = overbrace(sqrt(K_a))^(color(blue)("constant")) * 1/sqrt(c)#

Since **constant** at a given temperature,

#color(green)(beta prop 1/sqrt(c))#

Indeed, the degree of dissociation is **inversely proportional** to the square root of

#c uarr implies 1/sqrt(c) darr" "# and#" " c darr implies 1/sqrt(c) uarr#