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Sulfuric acid donates 2 equiv of $H_3O^+$ to aqueous solution.

$H_2SO_4 + H_2O rarr HSO_4^(-) + H_3O^+$

$HSO_4^(-) + H_2O rarr SO_4^(2-) + H_3O^+$

The first equilibrium lies almost quantitatively to the right. The second equilibrium is smaller, but still operates, $K_(a2)=1.2xx10^-2$.

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