Question

Why is Ka1 = Ka2 for H2SO4 in water ?​

Answer 1

This is because a neutral H2SO4 has a much higher tendency to lose a proton than the negatively charged HSO4- . Thus, the former is a much stronger acid than the latter.

Answer 2

Ka1 and Ka2 are the first and second ionization constants of H2SO4, respectively. The dissociation of H2SO4 in water can be represented by the following equations:

$H_{2}$S$O_{4}$ ↔ $H^{+}$ + HS$O_{4} ^{-}$- (Ka1)

HS$O_{4}$- ↔ $H^{+}$ + S$O_{4}^{2-}$- (Ka2)

• The value of Ka1 for $H_{2}$S$O_{4}$  is much larger than that of Ka2, indicating that the first ionization is more favorable than the second one. This can be explained by the following reasons:
• The first proton is easier to remove: The first hydrogen ion is held more loosely by the sulfate ion due to its greater distance from the negatively charged oxygen atoms. Therefore, it is easier to remove the first proton from $H_{2}$S$O_{4}$  than the second one.
• The second proton is strongly held: Once the first proton is removed, the remaining $H_{2}$S$O_{4} ^{-}$  ion becomes negatively charged, making it more difficult to remove the second hydrogen ion. The sulfate ion also becomes more negatively charged, further strengthening the bond with the hydrogen ion.
• The concentration of H+ ions: As the first proton is removed from  $H_{2}$S$O_{4}$ , the concentration of H+ ions in the solution increases, making it more difficult to remove the second proton.
• Therefore, we can conclude that Ka1 is much larger than Ka2 for  $H_{2}$S$O_{4}$  in water. However, it is worth noting that at very high acid concentrations, the dissociation behavior may deviate from what is expected due to the influence of interionic interactions. Nonetheless, in dilute solutions of  $H_{2}$S$O_{4}$ , Ka1 is essentially equal to Ka2, as the deviation from ideal behavior is negligible in these conditions.

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