This is theoretical scheme for preparing $\ce{KHSO4}$: $$\ce{K2CO3 + 2H2SO4 -> 2KHSO4 + CO2 +H2O}$$ using 1 eq $\ce{K2CO3}$ and 2 eq of $\ce{H2SO4}$ in approx 10 eq of $\ce{H2O}$ one gets $\ce{K2SO4}$ instead. How is it possible that $\ce{H2SO4}$ completely dissociated despite $\ce{HSO4-}$ is very weak acid in water? Does $\ce{K+}$ ion somehow influence the acidity of that anion? You have to use double the amount of sulfuric acid (4 eq) to get desired product. Also when using 1 eq $\ce{K2CO3}$ and 1 eq of $\ce{H2SO4}$ what product would you expect? Mixture of $\ce{KHCO3}$ and $\ce{KHSO4}$?


1 Answer 1


The pKa2 of sulfuric acid is 1.99 (Table of pka). This is definitely less acidic than pKa1, but calling this a 'weak' acid may be a distraction.

The relevant question is: does $\ce{KHCO3}$ readily deprotonate $\ce{KHSO4}$? The pKa1 of carbonic acid is 6.35 (same source), making it a weaker acid than $\ce{KHSO4}$. The difference is more than 4 orders of magnitude, so yes, I expect that $\ce{KHSO4}$ will more strongly favor depronotation than carbonic acid, and hence $\ce{K2SO4}$ is the product I expect.

To get at this quantitatively, consider the sub reaction:

$$\ce{KHCO3 + KHSO4 <=> K2SO4 + H2CO3 }$$

Which leads to the expression: $$\ce{K_{eq} = \frac{[K2SO4][H2CO3]}{[KHCO3][KHSO4]} }$$ which can be rewritten as $$\ce{K_{eq} = \frac{[H^+][K2SO4][H2CO3]}{[H^+][KHCO3][KHSO4]} }$$ Which can be rewritten as $$\ce{K_{eq} = \frac{[H2CO3]}{[H^+][KHCO3]}\cdot\frac{[H^+][K2SO4]}{[KHSO4]}}$$ which can be written as $$\ce{K_{eq} = \frac{$K$_{a2,\ce{KHSO4}}}{$K$_{a1,\ce{H2CO3}}} }$$ Taking the log of both sides $$\ce{pK_{eq} = p$K$_{a2,\ce{KHSO4}} - p$K$_{a1,\ce{H2CO3}} = } -4.36$$

Which is $$\ce{K_{eq} \approx 23,000}$$ heavily favoring products, and hence, $\ce{K2SO4}$ is the dominant product


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.