# What is the K_b for HSO_4^(-) if its K_a is 1.99?

Aug 2, 2017

This question can't quite be answered in its exact wording...

...because only the ${K}_{a 1}$ of ${\text{H"_2"SO}}_{4}$ is related to the ${K}_{b}$ for ${\text{HSO}}_{4}^{-}$, and only the ${K}_{a}$ of ${\text{HSO}}_{4}^{-}$ (i.e. the ${K}_{a 2}$ of ${\text{H"_2"SO}}_{4}$) is related to the ${K}_{b}$ of ${\text{SO}}_{4}^{2 -}$.

${\text{H"_2"SO"_4 stackrel(K_(a1)" ")(rightleftharpoons) "HSO"_4^(-) stackrel(K_(a2)" ")(rightleftharpoons) "SO}}_{4}^{2 -}$
${\text{ "" "" "^(K_(b1))" "" "" "" }}^{{K}_{b 2}}$

Answering the question as-written would require looking up the ${K}_{a 1}$ of ${\text{H"_2"SO}}_{4}$, which is $\approx$ $1000$.

Recall that

${K}_{a} {K}_{b} = {K}_{w} = {10}^{- 14}$,

${\text{pK"_a + "pK"_b = "pK}}_{w} = 14$,

at ${25}^{\circ} \text{C}$ and $\text{1 atm}$.

So, the "${K}_{b}$ of ${\text{HSO}}_{4}^{-}$" (as you have stated) is...

${K}_{b 1} = {K}_{w} / {K}_{a 1} = {10}^{- 14} / \left({10}^{3}\right) \approx \underline{{10}^{- 11}}$

However, what you probably meant was the ${K}_{b}$ of $\underline{{\text{SO}}_{4}^{2 -}}$, the conjugate base of ${\text{HSO}}_{4}^{-}$... but another issue is that I don't believe your ${K}_{a}$.

The ${K}_{a}$ of ${\text{HSO}}_{4}^{-}$ is about $0.012$, so you have supplied the ${\text{pK}}_{a}$, i.e. you have given

${\text{pK}}_{a} = - \log \left({K}_{a}\right)$

Thus...

${10}^{- {\text{pK}}_{a}} = {K}_{a} = {10}^{- 1.99} = 0.0102$ is your ${K}_{a}$.

And so, the ${K}_{b}$ for ${\text{SO}}_{4}^{2 -}$ (which is probably what you actually thought you wanted) is...

$\textcolor{b l u e}{{K}_{b 2}} = {K}_{w} / {K}_{a 2}$

$= {10}^{- 14} / 0.0102$

$= \underline{\textcolor{b l u e}{9.77 \times {10}^{- 13}}}$

With a ${K}_{b}$ this small, is ${\text{HSO}}_{4}^{-}$ primarily an acid that shall dissociate, or a base that shall associate? i.e. will it be easier to form sulfate, or sulfuric acid in aqueous solution?