How to find the ionization constant of an unknown weak acid? A $\pu{0.45 M}$ solution of a weak acid, $\ce{HX}$, has a $\pu{pH}$ of $4.5$.  What is the ionization constant, $K_a$, of the acid?



$\ce{[HX]}$ is already given as $\pu{0.45 M}$.
$\ce{[H+]}$ is given by $10^{-4.5}$.


Since $\ce{pH}$ is $4.5$, $\ce{pOH} = 14 - 4.5 = 9.5$. Then we can get $\ce{[OH^-]}$ by computing $10^{-9.5}$. 

Then computing $K_a$, I get 
$$\frac{10^{-4.5} \cdot 10^{-9.5}}{0.45} = 2.22 \times 10^{-14}.$$
But that is apparently not the right answer. What did I do wrong?
Look at the answer given here and modify it:
How to set up equation for buffer reaction?

In your case the concentration of base $c_B \approx 0$ and $[\ce{OH^-}] = K_\mathrm{w}/[\ce{H^+}] \approx 0$, so that the equation reduces to $$K_\mathrm{a} = \frac{[\ce{H^+}]^2}{c_\mathrm{a}-[\ce{H^+}]} \approx \frac{[\ce{H^+}]^2}{c_\mathrm{a}},$$ 
as $c_\mathrm{a}$ is so large.