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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?

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

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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:

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

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  • $\begingroup$ 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. $\endgroup$ – Martin - マーチン Nov 4 at 11:10
  • $\begingroup$ 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. $\endgroup$ – James Gaidis Nov 4 at 14:25

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