$$\begin{align} \ce{H2O &<=> H+ + OH-} \\ K &= \frac{[\ce{H+}][\ce{OH-}]}{[\ce{H2O}]} \\ \ce{H2SO4 &<=> 2H+ + SO4^2-} \end{align}$$
In acidulated water water dissociates into its respective ions(though the dissociation is very less) and simultaneously acid dissociates into its respective ions.
Now as the acid (I have taken $\ce{H2SO4}$ here) also dissociates to give $\ce{H+}$ ions. So the $\ce{H+}$ ion concentration increases which in turn affects the equilibrium of $\ce{H2O}$ and favours the backward reaction forming $\ce{H2O}$ back. So can we say common ion effect has suppressed the dissociation of $\ce{H2O}$?
And if this is the case then why do we use acidulated water while performing the electrolysis of water? (I mean the common ion effect can suppress dissociation of water and we would get less $\ce{H+}$ and $\ce{OH-}$ ions in the medium which would lead to a lower yield if $\ce{H2}$ and $\ce{O2}$ gas.)
Answer
You have a couple of concepts mixed together in the wrong way.
(1) The ionization of water is typically written as
$\text{K}_\text{w} = \ce{[H+] \cdot [OH^-]}$
The assumption is that the ionization is so low that the concentration of water remains a constant. For a solution which is half water and half acetic acid that wouldn't be true of course.
(2) The dissociation and recombination of water molecules is a very fast reaction and occurs continuously in water.
$\ce{2H2O <=> H3O+ + OH-}$
(3) In reality the autoprotolysis of water is dependent on the activities of the $\ce{H+}$ and $\ce{OH-}$. So the relation is more properly written as
$\text{K}_\text{w} = a_\ce{H+} \cdot a_\ce{OH^-}$
In dilute solutions the activity is essentially the same as the concentration, but in more concentrated solutions the activity drops as the concentration increases because of the formation of clusters of ion pairs in solution.
Also the activity drops as the overall ionic strength of the solution increases for the same reason. So if you add $\ce{NaCl}$ to the solution, then the ionic strength of the solution increases and the activity of the hydronium ions and hydroxide ions decreases. So in concentrated $\ce{NaCl}$ the $\ce{Na+}$ and $\ce{Cl-}$ ions aren't really just "spectators."
(4) If you add $\ce{H2SO4}$ to distilled water then the acid will dissociate according to the following reactions:
$\ce{H2SO4 + H2O <=> H3O+ + HSO4-}\quad \quad \text{pK}_{\alpha1} = -3$
$\ce{HSO4- + H2O <=> H3O+ + SO4^{2-}}\quad \quad \text{pK}_{\alpha2} = 1.99$
The final pH will depend on the final concentration of the sulfate species which is commonly expressed as the molarity of the sulfuric acid. So if we diluting concentrated sulfuric acid we typically use
$\text{m}_1 \cdot \text{V}_1 = \text{m}_2 \cdot \text{V}_2$
where $\text{m}$ is the molarity of the solution and $\text{V}$ is the volume of the solution. This ignores the fact that in dilute solution the sulfuric acid isn't just $\ce{[H2SO4]}$ but that the overall sulfuric acid concentration is the sum of $\ce{[H2SO4]}$, $\ce{[HSO4^-]}$, and $\ce{[SO4^{2-}]}$.
(5) Acidifying an electrolysis solution isn't really about ion concentration since adding NaCl isn't as effective as adding HCl. The gist is that protons can move through solution very quickly since a water molecule can accept a proton from one direction to form hydronium and kick off a different proton in the other direction to form water again. So a net charge can move through solution quickly as a result of the chain reaction of such transfers. Thus any individual proton doesn't have to travel far.
CONCLUSION
The overall notion would be better expressed as that adding sulfuric acid to water increases the $[\ce{H+}]$ and reduces the $[\ce{OH-}]$ thus shifting the equilibrium for the autoprotolysis of water.
No comments:
Post a Comment