# Question 69f70

Aug 2, 2015

As physical scientists we should look at the numbers; $H S {O}_{4}^{-}$ ($p {K}_{a}$ = 1.92) is properly described as a moderately strong acid.

#### Explanation:

The $p {K}_{a}$ of $H S {O}_{4}^{-}$ is 1.92. I will let you do the calculation for say a 1.0 $m o l {L}^{-} 1$ solution (you will probably have to solve it exactly!). Its sodium salt is commonly called "dry acid" in the outdoor swimming pool industry.

Aug 2, 2015

Because ${H}_{2} S {O}_{4}$ is a strong acid.

#### Explanation:

Hydrogen sulfate, HSO_4""^(-)#, results from the dissociation of sulfuric acid, ${H}_{2} S {O}_{4}$, in aqueous solution.

${H}_{2} S {O}_{4 \left(a q\right)} + {H}_{2} {O}_{\left(l\right)} r i g h t \le f t h a r p \infty n s {H}_{3} {O}_{\left(a q\right)}^{+} + H S {O}_{4 \left(a q\right)}^{-}$

The very important thing to realize here is that this equilibrium lies so far to the right that this reaction can be considered as going to completion.

${H}_{2} S {O}_{4 \left(a q\right)} + {H}_{2} {O}_{\left(l\right)} \to {H}_{3} {O}_{\left(a q\right)}^{+} + H S {O}_{4 \left(a q\right)}^{-}$

This means that the resulting solution will contain no ${H}_{2} S {O}_{4}$ molecules. Sulfuric acid's strength as an acid comes from the fact that it will dissociate completely in aqueous solution.

So, if you add sulfuric acid to water, you can expect all the sulfuric acid molecules to donate one proton and dissociate into hydrogen sulfate and hydronium.

This implies that the hydrogen sulfate anion simply cannot accept a proton, i.e. act as a base, to reform sulfuric acid because the aforementioned equilibrium will not allow for the reverse reaction to take place.

$H S {O}_{4 \left(a q\right)}^{-} + {H}_{2} {O}_{\left(l\right)} \textcolor{red}{\cancel{\textcolor{b l a c k}{\to}}} {H}_{2} S {O}_{4 \left(a q\right)} + O {H}_{\left(a q\right)}^{-}$