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What happen when Cl(2) is passed through...

What happen when `Cl_(2)` is passed through a hot concentrated solution of a base like `Ba(OH)_(2)`?

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On treatment with `Ba(OH)_(2), Cl_(2)` undergoes disproportionation to form barium choride, `BaCl_(2)` and barium chlorate, `Ba(ClO_(3))_(2)`.
`6 Ba(OH)_(2)(conc.)+6 overset(0)(Cl_(2))overset("Hot") rarr 5 overset(-1)(BaCl_(2))+overset(+5)(Ba(ClO_(3))_(2))+6 H_(2)O`
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