12
$\begingroup$

In my textbook, it is given that the ionic product of water $K_\mathrm{w}$ remains constant even when a non-neutral solute such as an acid is added to it.

$$K_\mathrm{w} = \ce{[H3O+][OH-]}$$

When a strong acid is added to water, the concentration of $\ce{[H3O+]}$ increases. Yet the $K_\mathrm{w}$ must remain constant. But how? How does the concentration of $\ce{[OH-]}$ decrease?

Improve this question
$\endgroup$

2 Answers 2

Reset to default
8
$\begingroup$

$K_w$ is the equilibrium constant of this reaction: $$\ce{2H2 O <=> H3O+ +OH-}.$$ So, it's constant at a fixed temperature. It equals $10^{-14}$ at $25~^\circ\mathrm{C}$. At this temperature $\ce{[H3O+]=[OH^{-}]}=10^{-7}~\mathrm{mol/L}$.

If you add acid to water, the above reaction equilibrium is displaced in order to counteract the change in the concentration of ion hydronium, i.e. the reaction is displaced to the left. This means that ions hydroxide will react with the excess of ions hydronium to give water. As a result: the concentration of ion hydroxide will be less than $10^{-7}~\mathrm{mol/L}$, and the concentration of ion hydronium will be more than $ 10^{-7}~\mathrm{mol/L}$. But their product is constant. $\ce{[H3O+] \cdot[OH^{-}]}=10^{-14}$.

Improve this answer
$\endgroup$
2
  • 1
    $\begingroup$ If excess kfhydronium reacted with hydroxide , won't conc. of both decrease. I.e. H+ conc. still remains $10^{-7}$ ? $\endgroup$
    – Anubhav Goel
    Nov 3, 2016 at 14:24
  • 1
    $\begingroup$ This explains what happens but not why. $\endgroup$
    – lukejanicke
    May 16, 2019 at 15:22
2
$\begingroup$

In Layman’s views, you can also think of it this way: initially, you have water, a small amount of $\ce{H3O+}$ and a small amount of $\ce{OH-}$. If you add an external proton source, we can consider what will happen to each of these:

\begin{array}{lccc}\hline \text{time point} & \text{acid} & \text{neutral} & \text{base} \\ \hline \text{before a.} & \ce{H3O+} & \ce{H2O} & \ce{OH-} \\ \text{after a.} & \text{no change} & \ce{H3O+} & \ce{H2O} \\ \hline \end{array}

In words: Nothing will happen to $\ce{H3O+}$ — at least nothing notable; $\ce{H4O^2+}$ is too unstable in aquaeous solution to contribute meaningly. Water will be protonated to $\ce{H3O+}$ and hydroxide ions will also be protonated (statistically) by the incoming acid to give $\ce{H2O}$. Thus, you can consider the addition of acid to both increase oxonium ion concentration and reduce hydroxide ion concentration.

The equilibrium constant and equation is just a more formal and scientifically more correct way of putting this.

Improve this answer
$\endgroup$
3
  • $\begingroup$ This was helpful! But I am still looking for an explanation of why the increase in $\ce{H3O+}$ and decrease in $\ce{OH−}$ happens in just such as way that the product of their concentrations remains constant, instead of… say, the sum of their concentrations remaining constant, which is what I intuitively thought should happen after looking at your table. $\endgroup$
    – lukejanicke
    May 16, 2019 at 15:07
  • $\begingroup$ @lukejanicke That basically books down to physical chemistry and equilibrium theory whose formulae I know how to apply but not how to derive from scratch. Sorry … $\endgroup$
    – Jan
    May 17, 2019 at 2:53
  • 1
    $\begingroup$ I haven’t been able to find an answer yet. Someone asked this and was told the question is a duplicate of this, which I disagree with, but the answer might be in there somewhere. $\endgroup$
    – lukejanicke
    May 17, 2019 at 4:01

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.