# How do we assess the strength of an acid?

Jul 27, 2017

Normally, we consider an equilibrium reaction, and must specify a SOLVENT, which is typically water.......

#### Explanation:

We represent the general reaction of an acid in water by the following equilibrium.......

$H X \left(g\right) + {H}_{2} O \left(l\right) r i g h t \le f t h a r p \infty n s {H}_{3} {O}^{+} + {X}^{-}$

The extent to which this equilibrium lies to the RIGHT, specifies the STRENGTH of the acids. Acids such as ${H}_{2} S {O}_{4} , H C l {O}_{4} , H X \left(X \ne F\right)$, are all strong acids. And this ${H}_{3} {O}^{+}$, the hydronium ion, is CONCEIVED to be a protonated water molecule.........i.e. ${H}^{+} + {H}_{2} O \rightarrow {H}_{3} {O}^{+}$ (mass and charge are conserved as always!) The $H X$ molecule has been IONIZED, and we get charge-separated ions, formally ${H}^{+}$ or ${X}^{-}$.

And thus $H C l \left(g\right)$ (for instance) is a source of hydronium ion, ${H}_{3} {O}^{+}$ in aqueous solution.........

We may take a tank of $H C l \left(g\right)$, a room temperature gas, and we can bleed it in to water to give an AQUEOUS solution that we could represent as $H C l \left(a q\right)$ OR ${H}_{3} {O}^{+}$ and Cl^−.

$H C l \left(g\right) \stackrel{{H}_{2} O}{\rightarrow} {\underbrace{{H}_{3} {O}^{+}}}_{\text{hydronium ion}} + C {l}^{-}$

In each case this is a REPRESENTATION of what occurs in solution. If we bleed enuff gas in, we achieve saturation at a concentration of approx. $10.6 \cdot m o l \cdot {L}^{-} 1$ with respect to hydrochloric acid - of course the acid has been $\text{ionized}$.

As far as anyone knows, the actual acidium ion in solution is ${H}_{3} {O}^{+}$, or ${H}_{5} {O}_{2}^{+}$, or ${H}_{7} {O}_{3}^{+}$, i.e. a cluster of 2 or 3 or 4 water molecules with an EXTRA ${H}^{+}$ tacked on. We represent it in solution (without loss of generality) as ${H}_{3} {O}^{+}$, the $\text{hydronium ion}$, which is clearly the conjugate acid of ${H}_{2} O$. Sometimes, especially when we are first introduced to the concept, we represent it simply as the $\text{protium ion}$, i.e. ${H}^{+}$, which is perhaps a bit simpler, but ignores the water solvent.

Note that the ${H}^{+}$ is quite mobile, and passes, tunnels if you like, the extra ${H}^{+}$ from cluster to cluster. If you have ever played rugby, I have always liked to compare this to when the forwards form a maul, and can pass the pill from hand to hand to the back of the maul while the maul is still formed.

Of course, tunnelling, proton transfer, is more likely in a cluster of water molecules, so the analogy might not be particularly apt in that there is definite transfer of a ball in a maul, but a charge in a water cluster is conceivably tunnelled. The same applies to the transfer of an hydroxide ion. For this reason both ${H}^{+} \text{/} {H}_{3} {O}^{+}$ and $H {O}^{-}$ have substantial mobility in aqueous solution.

And we know that hydroxide/hydronium ions obey the following equilibrium in aqueous solution at $298 \cdot K$.......

${H}_{3} {O}^{+} + H {O}^{-} \rightarrow 2 {H}_{2} O$; ${K}_{w} = \left[{H}_{3} {O}^{+}\right] \left[H {O}^{-}\right] = {10}^{- 14}$.

Depending at which level you are at (and I don't know!, which is part of the problem in answering questions on this site), you might not have to know the details at this level of sophistication. The level I have addressed here is probably 1st/2nd year undergrad.........