# What is the ionization constant, $K_a$, of the acid?

A 0.45 M solution of a weak acid, HX, has a pH of 4.5. What is the ionization constant, $K_a$, of the acid?

Attempt at solution:

[HX] is already given as 0.45 M.

[H+] is given by $10^{-4.5}$.

Since pH is 4.5, pOH $= 14 - 4.5 = 9.5$.Then we can get $[OH^{-}]$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?

• $$K_\mathrm a = \frac{[\ce{H+}][\ce{X-}]}{[\ce{HX}]} \neq \frac{[\ce{H+}][\ce{OH-}]}{[\ce{HX}]}$$ – orthocresol Apr 22 '18 at 13:38

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 $\ce{[OH^-]=Kw/[H^+] \approx 0}$ so that the equation reduces to $K_A=\ce{[H^+]^2/(c_a-[H^+]) \approx [H^+]^2/(c_a)}$ as $c_a$ is so large.
$$K_{a} = \frac{[H^{+}] [A^{-}]} {[HA]}$$
In terms of data, [A-] hasn't been provided, and so we can assume that [H+][A-] $\approx$ [H+]2. Also, the [H+] you provided looks more like a pH conversion, so [H+] = 10-4.5 = 3.16x10-5 moldm-3. Putting all of this together means that:
$K_{a} \approx \frac{[H^{+}]{^{2}}} {HA}$ = $\frac{(3.16*10^{-5}){^{2}}} {0.45}$ = 2.22x10-9 moldm-3.