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On the reaction of syn-dihydroxylation with stoichiometric quantities of OsO4 (not a catalytic amount) the second step is $\ce{NaHSO3/H2O}$ or $\ce{Na2SO3/H2O}$.

According to my understanding, $\ce{H2O}$ is the only thing needed to hydrolyze the cyclic osmate ester resulting from the first step of adding $\ce{OsO4}$.

So, what is the use of $\ce{NaHSO3}$ or $\ce{Na2SO3}$ here?

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You need to reduce remaining oxidizer (OsO4) and maintain pH of the solution. NaHSO3 acts both as the reducing agent and a buffer.

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