In adding $\ce{^{16}O}$-sodium hydroxide to $\ce{^{18}O}$-water to get a $0.25\pu M$ solution, is the hydroxide in form of $\ce{^{16}O}$-hydroxide or $\ce{^{18}O}$-hydroxide?

is there an oxygen exchange between $\ce{^{16}O}$-hydroxide and $\ce{^{18}O}$-water?

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    $\begingroup$ Yes they will exchange, and pretty fast, too. $\endgroup$ – Ivan Neretin Aug 16 '18 at 15:34

Yes, it will but it will be a hydrogen exchange, not an oxygen exchange. $$\ce{2 H2^{18}O <=> \color{blue}{H3^{18}O+} + ^{18}OH-}$$ $$\ce{\color{blue}{H3^{18}O+} + ^{16}OH- <=> H2^{18}O + H2^{16}O }$$ The oxygen will mix with water and hydroxide to a near equivalent ratio of $\ce{^{18}O:^{16}O}$ in both $\ce{H2O}$ and $\ce{OH-}$ but not perfectly equivalent as isotopes do not have equivalent chemical potentials.

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  • $\begingroup$ if only a bit of 16OH- was added into the 18O-water, can I assume that your chemical equations go to the right side infinitely? finally vast majority of the hydroxides are in form of 18OH-?? thank you.. could you please provide a link of a reference? $\endgroup$ – Liming Wang Aug 20 '18 at 8:13
  • $\begingroup$ @LimingWang The reactions are reversible. Reference: autodissociation of water $\endgroup$ – A.K. Aug 20 '18 at 14:26

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