0
$\begingroup$

I am studying $K_a$ currently and there is a passage in the book that doesn't appear to make sense.

It states that from the equation

$$\ce{HA <=> H+ + A-}$$

that $$K_a = \frac{[H^+][A^-]}{[HA]}$$

The following passage says When dealing with weak acids, you can assume that dissociation of the acid is much greater than the dissociation of water.

It continues to say that you can assume all the H+ ions come from the acid.

hence $$K_a = \frac{[H^+]^2}{[HA]}$$

What I don't understand is the dissociation of water.

The book gives the equation of dissociation of water as the following -

$$\ce{H_2O(l) + H_2O(l) <=> H3O+ + OH-}$$

but where does the H+ come from from this dissociation - it's made a Hydroxonium ion?

Is it because the $H_3O^+$ can act as the conjugate acid and release the proton?

Please help!

$\endgroup$
0
2
$\begingroup$

Yes, water autodissociates according to the equation $\ce{2H2O <=> H3O+ + OH-}$.

But I should point out that the equilibrium constant is written as though the concentration of water itself is a constant, so the reactions becomes:

$\ce{H2O <=> H+ + OH-}$.

and the equilibrium is expressed as:

$K_\rm{w} = 1\times10^{-14} = \ce{[H+][OH-]}$

The pH is $\text{-log}\ce{([H+])}$ thus the pH of pure water is 7.

This also has to do with significant figures. Few pKa's are known to better than two significant figures. So:

  • For weak acid with pKa < 5 you can ignore water contribution.

  • For acid with pKa > 9 you can assume all the $\ce{H+}$ is from water and that all of the acid is essentially present as the HA form.

  • For 9 > pKa > 5 you have to consider both ionization of the acid and water.

$\endgroup$
2
$\begingroup$

I think MaxW's answer covers the gist of the question from the calculation side, but its also worth noting something else about your question. $\ce{H^+}$ and $\ce{H3O^+}$ are often used interchangeably for the acidic proton, but $\ce{H3O^+}$ is more accurate; free protons are unlikely to exist in solution. $\ce{H3O^+}$ is one example of how the proton could be passed into the solution and so is used to represent the general case of the proton being solvated by a water cluster. For more information on these water clusters, go the solvation section of https://en.wikipedia.org/wiki/Hydronium.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.