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$$\begin{align} \ce{H2O &<=> H+ + OH-} \\ K &= \frac{[\ce{H+}][\ce{OH-}]}{[\ce{H2O}]} \\ \ce{H2SO4 &<=> 2H+ + SO4^2-} \end{align}$$

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.)

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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.

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  • $\begingroup$ (thus shifting the equilibrium for the autoprotolysis of water) toward dissociation or recombination of the water molecules. $\endgroup$ – Adnan AL-Amleh Jul 27 '18 at 18:06
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If you added pure, isolated $\ce{H+}$ (somehow) to water, it would disturb the equilibrium of the auto-ionization of water quite substantially.

However, you have to consider that in addition to $\ce{H+}$, the dissociation of sulfuric acid produces $\ce{SO4^{2-}}$ as you've written. This counter-ion will pull your first equilibrium to the right during electrolysis -- not by addition of $\ce{H2O}$, but by electrostatic attraction of $\ce{SO4^{2-}}$ to the hydrogen in $\ce{H2O}$.

In this regard, addition of sulfuric or other acids for use in electrolysis of water generally helps because the "mess" of ions serves to smooth the process of splitting $\ce{H2O}$. The net effect actually lowers the energy required to break $\ce{H2O}$. The same principle is useful for electroplating; I use acetic acid as an enhancement to ion-flow for plating metals.

Ionic strength is the quantitative effect of interest here:

$$ I = \frac{1}{2} \sum_i c_i z^2_i $$

Where $i$ are different ions, $c_i$ are concentrations of those ions, and $z_i$ are charges of those ions. In general, more ions in aqueous solution actually improves rate of electrolysis of water, or:

$$k_{\text{electrolysis}} \propto I$$

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  • $\begingroup$ Glad to help. Please accept by clicking the checkmark if you feel your question was answered $\endgroup$ – khaverim Jan 28 '17 at 7:12
  • $\begingroup$ I have serious doubts about this answer. For instance if you add HCl to water, then the assumption is that all the HCl dissociates. So the $\ce{Cl^-}$ is regarded as a "spectator" ion in that it doesn't influence the pH. Also I could dump in NaCl and the pH stays the same. // The rest of the story is that the equilibrium really depends on the activities of the ions which vary with the ionic strength of the solution. "Low" ionic concentrations are typically assumed and the concentration is used instead of the activity. $\endgroup$ – MaxW Jan 28 '17 at 20:33
  • $\begingroup$ @MaxW electrostatics have predominance over pH for electrolysis of water. I added "during electrolysis" to my answer for clarity. In fact, pH is practically irrelevant compared to ion concentration, current, voltage, electrode surface, etc. The distinction between electrolysis and auto-ionization is important because the OP asked for it $\endgroup$ – khaverim Jan 29 '17 at 5:13

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