# 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?

• $$K_\mathrm a = \frac{[\ce{H+}][\ce{X-}]}{[\ce{HX}]} \neq \frac{[\ce{H+}][\ce{OH-}]}{[\ce{HX}]}$$ – 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 $$[\ce{OH^-}] = K_\mathrm{w}/[\ce{H^+}] \approx 0$$, 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.

Forget about $$\ce{[OH^-]}$$. This is $$K_\mathrm{a}$$ we're talking about - the acid dissociation constant. Remember that the equation for $$K_\mathrm{a}$$ is:

$$K_\mathrm{a} = \frac{[\ce{H^{+}}] [\ce{A^{-}}]} {[\ce{HA}]}$$

As you can see, $$\ce{[OH^{-}]}$$ doesn't come into the equation.

In terms of data, $$\ce{[A^{-}]}$$ hasn't been provided, and so we can assume that $$[\ce{H^+}][\ce{A^-}] \approx [\ce{H^{+}}]^2.$$ Also, the $$\ce{[H^{+}]}$$ you provided looks more like a $$\ce{pH}$$ conversion, so $$[\ce{H^+}] = 10^{-4.5} = \pu{3.16x10^-5 mol dm^-3}$$. Putting all of this together means that:

$$K_\mathrm{a} \approx \frac{[\ce{H^{+}}]^2}{\ce{HA}} = \frac{(3.16\times10^{-5})^2}{0.45} = \pu{2.22x10^-9 mol dm^-3}.$$

• It is evident from $\ce{HA <=> H+ + A-}$, that $c(\ce{H+})=c(\ce{A-})$, hence $[\ce{H^+}][\ce{A^-}] = [\ce{H^+}]^2$. Nevertheless, this answer is still not quite correct, as it includes the approximation $c(\ce{HA})-c(\ce{H+}) \approx c(\ce{HA})$, which should be proven to hold here. – Martin - マーチン Nov 4 '19 at 11:10
• But c(HA + A-) = 0.45 M. Since we say cH+ = cA- , then cHA = 0.45 - cA- = 0.45 - cH+ = 0.45 - 3.16 e-5, which is ~0.45. – James Gaidis Nov 4 '19 at 14:25