# Acid dissociation constant calculated from the ionization percentage

The question goes like this:

An acid (HX) is 25% dissociated in water. If the equilibrium concentration of HX is $$\pu{0.30 M}$$, calculate the value of $$K_\mathrm{a}$$.

So I tried as follows:

if I know that the ionization percentage $$\alpha = 25\%$$

$$\frac{25}{100} = \frac{[\ce{HX}]_\mathrm{equilibrium}}{[\ce{HX}]_\mathrm{initial}} = \frac{0.30}{[\ce{HX}]_\mathrm{initial}}$$ Thus: $$[\ce{HX}]_\mathrm{initial} = \pu{1.2 M}$$

With this initial concentration I could calculate the concentration variation: $$[\ce{HX}]_\mathrm{equilibrium} - [\ce{HX}]_\mathrm{initial} = -0.9$$

Thus, knowing that the initial concentration of $$[\ce{H+}]$$ and $$[\ce{X-}]$$ is $$0$$, in equilibrium their concentration would be:

$$0 - (-0.9)= 0.9$$

And finally replacing this in the relation to the acid dissociation constant:

$$K_\mathrm{a} = \frac{[\ce{H+}][\ce{X-}]}{[\ce{HX}]} = \frac{(0.9)(0.9)}{(0.30)} = 2.7$$

But unfortunately, this procedure is wrong and doesn't match the answer of the exercise. Could someone tell me what I'm missing? Or where did I go wrong?

• The error is in the first equation line. Read the task in hand more carefully. Aug 15 at 14:36

Suppose the ionization of the given weak acid is as follows: $$\ce{HX (aq) + H2O <=> H3O+ (aq) + X- (aq)} \tag1$$
Suppose intial concentration is $$c$$ and $$\alpha$$ amount of $$c$$ is ionozed at the equlibrium $$(\text{both in } \pu{mol L-1})$$. Thus, the equlibrium concentrations of $$\ce{HX, X-},$$ and $$\ce{H3O+}$$ are $$(c-\alpha)$$, $$\alpha$$, and $$\alpha$$, respectively. Since 25% if acid is ionized at the equlibrium: $$\alpha = 0.25c \tag2$$ Since the equlibrium concentration of $$\ce{HX},$$ is $$\pu{0.30 mol L-1}$$: $$c - \alpha = 0.0.30 \tag3$$
$$K_\mathrm{a} = \frac{[\ce{H3O+}][\ce{X-}]}{[\ce{HX}]} = \frac{\alpha\cdot\alpha}{c-\alpha}$$