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?

  • 2
    $\begingroup$ The error is in the first equation line. Read the task in hand more carefully. $\endgroup$
    – Poutnik
    Commented Aug 15, 2021 at 14:36

1 Answer 1


Although this is a homework question, OP's effert to solve it deserve some insight. As Poutnik correctly pointed out, the error is in the OP's first equation (due to not reading the task in hand more carefully, see Poutnik's comment elsewhere).

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% of acid is ionized at the equlibrium: $$\alpha = 0.25c \tag2$$ Since the equlibrium concentration of $\ce{HX},$ is $\pu{0.30 mol L-1}$: $$c - \alpha = 0.30 \tag3$$

Now you have two unknowns and two independent equations so that using your mathematical knowledge you can solve the problem. The acid dissociation constant is:

$$K_\mathrm{a} = \frac{[\ce{H3O+}][\ce{X-}]}{[\ce{HX}]} = \frac{\alpha\cdot\alpha}{c-\alpha} = \frac{\alpha^2}{0.30}$$


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