Is a pH value outside 0 - 14 possible? I asked my teacher who said: yes, it is, but very difficult to achieve.
Then on the internet, I found multiple answers, one saying it is but because of a fault in the pH glass meter we can't be sure, the other said that it is not possible because of the intrinsic value of water to ionize. As you can see, I'm getting mixed answers from different sources and I'm really confused now. Could somebody here give a concise answer whether or not it is possible? A deep argument will be valuable.
Answer
Let me add a bit more to the answers already given. As has been said, $\mathrm{pH}$ is nothing but a measure of the activity of protons ($\ce{H+}$) in a solvent - $\displaystyle\mathrm{pH} = -\log_{10} \ce{\ a_{H_{\text{solvated}}^{+}}}$. In dilute solutions, a solute's activity is approximately equal to its concentration, and so you can get away with saying $\mathrm{pH} = -\log_{10} \ce{[H_{solvated}^{+}]}$.
The "normal" aqueous $\mathrm{pH}$ scale that goes from 0-14 delimits the region where both $\ce{[H^{+}]}$ and $\ce{[OH^{-}]}$ are lower than $\pu{1 mol L^{-1}}$, which is about the upper limit where concentrations and activities of solutes in solution are approximately the same and they can be used interchangeably without introducing too much error.
(The reason this is true is because a solute in a very dilute solution behaves as if it surrounded by an infinite amount of solvent; there are so few solute species that they interact very little with each other on average. This view breaks down at high concentrations because now the solute species aren't far enough apart on average, and so the intermolecular interactions aren't quite the same.)
Not only is it rarer to talk about pH for very highly concentrated solutions, because activity has a much more subtle definition than concentration, but also because you have to produce quite concentrated solutions to get unusual values of pH. For example, very approximately, to get a litre of an aqueous solution of $\mathrm{pH} = -1$ using hydrogen perchlorate ($\ce{HClO4}$, a very strong monoprotic acid) you would need to dissolve around $\pu{1 kg}$ of $\ce{HClO4}$ in a few hundred $\pu{mL}$ of water. A litre of $\mathrm{pH} = 15$ aqueous solution using potassium hydroxide ($\ce{KOH}$) would also require several hundred grams of the hydroxide in about as much water. It's so much acid/base in a small amount of solvent that many acids/bases can't even dissolve well enough to reach the required concentrations.
There is also another way to come across unusual values of pH. We almost always talk about acids in bases in water, but they also exist in other media. Notice that in the definition of pH, no direct reference is made to water. It just happens that, for water, we have:
$$\ce{H2O (l) <=> H^+ (aq) + OH^{-} (aq)} \\ K^{\pu{25^\circ C}}_{\text{autodissociation}}=k^{\pu{25^\circ C}}_\mathrm{w}=a_{\ce{H^+(aq)}} \times a_{\ce{OH^- (aq)} } \simeq \ce{[H^{+}]} \times \ce{[OH^{-}]} = 1.01\times 10^{-14}$$ $$\mathrm{pH}+\mathrm{pOH}=-\log\ k_\mathrm{w} \simeq 14 \ (25^\circ C)$$
However, for liquid ammonia, one has:
$$\ce{NH3{(l)} <=> H^+{(am)} + NH2^{-}{(am)}} \\ K^{\pu{-50^\circ C}}_{\text{autodissociation}}=k^{\pu{-50^\circ C}}_\mathrm{am}=a_{\ce{H^+(am)}} \times a_{\ce{NH2^- (am)} } \simeq \ce{[H^{+}]} \times \ce{[NH2^{-}]} = 10^{-33}$$ $$\mathrm{pH}+\mathrm{p}\ce{NH2}=-\log\ k_{\mathrm{am}} \simeq 33 \ (-\pu{50^\circ C})$$
($\ce{(am)}$ stands for a substance solvated by ammonia). Hence, in liquid ammonia at $-50^\circ C$, the pH scale can easily go all the way from 0 to 33 (of course, it can go a little lower and higher still, but again now activities become important), and that neutral pH is actually 16.5.
For pure liquid hydrogen sulfate (some difficulties arise as $\ce{H2SO4}$ is a diprotic acid and tends to decompose itself when pure, but putting that aside and looking only at the first dissociation):
$$\ce{H2SO4{(l)} <=> H^+{(hs)} + HSO4^{-}{(hs)}}\\ K^{\pu{25^\circ C}}_{\text{autodissociation}}=k^{\pu{25^\circ C}}_\mathrm{hs}=a_{\ce{H^+(hs)}} \times a_{\ce{HSO4^- (hs)} } \simeq \ce{[H^{+}]} \times \ce{[HSO4-]} \simeq 10^{-3}$$ $$\mathrm{pH}+\mathrm{p}\ce{HSO4}=-\log\ k_{\mathrm{hs}} \simeq 3 \ (\pu{25^\circ C})$$
($\ce{(hs)}$ stands for a substance solvated by hydrogen sulfate). Thus, the pH in liquid hydrogen sulfate has a much smaller range than in water and ammonia, not going much further than interval from 0 to 3.
Notice also that not only does the range of $\mathrm{pH}$s change in each solvent, but there is no direct relationship between the values of pH between different solvents; formic acid in ammonia would behave as a strong acid and as such a $\pu{1 M}$ solution in ammonia would have a pH close to 0, but formic acid is actually a base in liquid hydrogen sulfate, and as such a $\pu{1 M}$ solution would have a pH above 1.5. An approximate comparison between the pH ranges and their relative positions in different solvents can be found in the figure below, from A Unified pH Scale for All Phases.
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