Chemistry - Is sulfuric acid or hydrochloric acid stronger?

Solution 1:

When using $\mathrm{p}K_\mathrm{a}$, one typically does not consider multiple dissociations for polyprotic acids, as the acidity of the conjugate base ($\ce{HSO4-}$ in the case of sulfuric acid) can and should be measured (or calculated) separately. Using that metric, $\ce{HCl}$ is stronger (at least per Wikipedia's acid strength page). One needs to keep in mind that $\mathrm{p}K_\mathrm{a}$ is solvent-dependent, and that the values typically given are for relatively dilute solutions. Acidity can sometimes vary widely with concentration ($\ce{HF}$ being a notorious example). It should also be noted that experimentally measuring acidity for very strong acids is actually quite difficult for numerous reasons.

On the other hand, if you evaluate acid strength by, say, the $\mathrm{pH}$ of the resulting solution, then you'd need to take subsequent dissociations into account. Using the crude approximation that an aqueous "strong acid" dissociates completely, then any diprotic "strong acid" is going to be stronger than a monoprotic one for equal concentrations (assuming a non-zero $K_\mathrm{a}$ for the second dissociation). Of course, this is not accurate, and a proper calculation would take into account not only $K_\mathrm{a}$ values for all dissociations, but also water's auto-ionization. I suspect that $\ce{HCl}$ would still be stronger, given that the $K_\mathrm{a}$ for the dissociation of $\ce{HSO4-}$ is fairly small, but one would need to actually crunch the numbers to be sure.

Solution 2:

Hydrochloric acid is the stronger of the two. It has a $\ce{pK_{a}}$ around -6.3 while the $\ce{pK_{a}}$ of sulfuric acid is only around -3. Here are the chemical equations for the dissociations of the two acids. You can see that the second dissociation in the case of sulfuric acid is not very extensive and doesn't contribute much to the acidity of sulfuric acid.

\begin{aligned} \ce{H2SO4 + H2O ~&<=>~ H3O+ + HSO4^{−}&&pK_{a}~=~-3}\\ \ce{HSO4- + H2O ~&<=>~ H3O+ + SO4^{−2}&&pK_{a}~=~2}\\ \ce{HCl + H2O ~&<=>~ H3O^+ + Cl^{-}&&pK_{a}~=~$-6.3$}\\ \end{aligned}

Solution 3:

While the other questions cover the issue fairly well, they omit one fairly important pragmatic reason for the strength of sulfuric acid over hydrochloric:

Sulfuric acid is a liquid, and may be almost pure. Hydrochloric acid is hydrogen chloride gas dissolved in water, with a maximum concentration of about 38%.

The extra water weakens the strength available for common use. For example, pure sulfuric acid will char sugar, while concentrated $\ce{HCl}$ will not.

Solution 4:

Although sulfuric acid has two acidic protons, only one of those protons dissociates completely in solution. After the first proton has dissociated, leaving $\ce{HSO4-}$, that species is now a weak acid as well as a weak base. As weak acids do not dissociate completely in solution, the remaining acidic proton is now "less" acidic.

A strong acid has a weak conjugate base. The $\ce{Cl-}$ ion created in solution by the dissociation of $\ce{HCl}$ is a terrible base, making hydrochloric acid very strong. The hydrogen sulfate anion is a weak base also, but it is a better base then a $\ce{Cl-}$ ion, adding to its weakened acidity.

While $K_\mathrm{a}$ values are a nice measure of acidity, only weak acids have a $K_\mathrm{a}$ value associated with them. Both hydrochloric acid and sulfuric acid are strong acids. In fact, because the $K_\mathrm{a}$ of sulfuric acid is so high, the eleventh edition (newest printed) textbook, Chang's "Chemistry", used in College chemistry, simply states it as "extremely high", assigning no numerical value to it. Likewise, the hydrogen sulfate anion is listed as having a $K_\mathrm{b}$ as "extremely low".