# 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?

Dec 22, 2015

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, ${\text{H"_3"O}}^{+}$, and the acid's conjugate base, will depend on the acid's dissociation constant, ${K}_{a}$.

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 $\text{HA}$ is said to be equal to $\beta$. This means that for every mole of $\text{HA}$ molecules, $\beta$ moles of each ion will be formed in solution.

For simplicity, I'll use hydrogen ions, ${\text{H}}^{+}$, instead of hydronium ions.

The dissociation of a weak acid looks like this

${\text{HA"_text((aq]) rightleftharpoons "H"_text((aq])^(+) + "A}}_{\textrm{\left(a q\right]}}^{-}$

As you know, the acid dissociation constant, ${K}_{a}$, is defined as

${K}_{a} = \left(\left[\text{H"^(+)] * ["A"^(-)])/(["HA}\right]\right)$

You know that

$\left[\text{HA}\right] = c$

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 $\text{HA}$ molecule that dissociates will produce $\beta$ ${\text{H}}^{+}$ and $\beta$ ${\text{A}}^{-}$ ions.

Plug this into the equation for ${K}_{a}$ to get

${K}_{a} = \frac{\beta \cdot \textcolor{red}{\cancel{\textcolor{b l a c k}{c}}} \cdot \beta \cdot c}{\textcolor{red}{\cancel{\textcolor{b l a c k}{c}}}} = {\beta}^{2} \cdot c$

Your goal now is to isolate $\beta$

${\beta}^{2} = {K}_{a} \cdot \frac{1}{c}$

Take the square root of both sides to get

$\beta = {\overbrace{\sqrt{{K}_{a}}}}^{\textcolor{b l u e}{\text{constant}}} \cdot \frac{1}{\sqrt{c}}$

Since ${K}_{a}$ is a constant at a given temperature, $\sqrt{{K}_{a}}$ will be a constant as well. This means that you have

$\textcolor{g r e e n}{\beta \propto \frac{1}{\sqrt{c}}}$

Indeed, the degree of dissociation is inversely proportional to the square root of $c$, since

$c \uparrow \implies \frac{1}{\sqrt{c}} \downarrow \text{ }$ and $\text{ } c \downarrow \implies \frac{1}{\sqrt{c}} \uparrow$