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If Ka2 is significantly less than Ka1, that means the concentration of products is less than that of the reactants for the Ka2 compared to Ka1. In other words, the second deprotonation of the acid is not favored and thus unlikely to occur.
To add on, it's because once you lose one proton, it gets much harder for another Hydrogen atom to leave when the molecule is now negatively charged. Because of this, the Ka will be smaller as it favors the reactants. However, H2SO4 is an exception where the first is fully deprotonized and is a strong acid in its first deprotonation.
Generally there is a trend, because as joannali1027 mentioned, after removing one proton (H+) from a polyprotic acid, the molecule has a negative charge (ex. H2PO4- from H3PO4 losing 1 proton). Negative charges attract positive charges, so the negatively charged molecule is more unlikely to let go of another, positively charged proton. So further deprotonation (product formation) is not favored, and Ka2 would be smaller since [products]<[reactants].
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