# Preparation of KHSO4

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 $\ce{KHSO4}$?

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.
$$\ce{KHCO3 + KHSO4 <=> K2SO4 + H2CO3 }$$
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 $$\ce{pK_{eq} = pK_{a2,\ce{KHSO4}} - pK_{a1,\ce{H2CO3}} = } -4.36$$
Which is $$\ce{K_{eq} \approx 23,000}$$ heavily favoring products, and hence, $\ce{K2SO4}$ is the dominant product