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?