If you compare the diffusion coefficient of hydroxide ($\pu{5.270e9 m^2/s}$) to that of fluoride ($\pu{1.460e9 m^2/s}$), you might be surprised to see such a difference despite their comparable size. The mechanism for the higher diffusion rate are acid/base reactions with neighboring waters. Instead of swapping positions, a water donates a proton to hydroxide, turning into hydroxide.

$$\ce{H-O- + H-O-H <=> H-O-H + O-H-}$$

With one deuterated water, it leads to formation of the desired product:

$$\ce{H-O- + D-O-D <=> H-O-D + O-D-}$$

If the acid/base reaction were slow compared to diffusion, there would be little effect on the diffusion constant of hydroxide, and it would be similar to that of fluoride. However, because the main mechanism of hydroxide transport is by acid/base (i.e. the protons move and the oxygen atoms don't have to), you get a more than four-fold difference. This means acid/base reactions are on a time scale faster than diffusion.

This is also reflected in the high molar conductivities, as stated in the comment by Poutnik.

So as stated in other answers, the exchange will be fast compared to the time required to mix the two liquids.

The analogous mechanism would work for $\ce{H3O+}$ but it is harder to compare to other ions, so I started with hydroxide. You can imagine that the rates are heavily dependent on temperature and pH. However, even at neutral pH the rate is fast.

Does some "easy" way exist to estimate the reaction speed in such systems, or it has to be determined experimentally?

Here is a paper where they used computational chemistry to look at this question: http://omh.umeche.maine.edu/pdfs/JChemPhys_135_124505.01pdf.pdf. I would not consider this "easy", though.

[Comment by Poutnik] Important is also en.wikipedia.org/wiki/Grotthuss_mechanism for the proton interchange. Mobility of H3O+ and OH- gives a hint it must be fast.

If you compare the diffusion coefficient of hydroxide ($\pu{5.270e9 m^2/s}$) to that of fluoride ($\pu{1.460e9 m^2/s}$), you might be surprised to see such a difference despite their comparable size. The mechanism for the higher diffusion rate are acid/base reactions with neighboring waters. Instead of swapping positions, a water donates a proton to hydroxide, turning into hydroxide.

$$\ce{H-O- + H-O-H <=> H-O-H + O-H-}$$

With one deuterated water, it leads to formation of the desired product:

$$\ce{H-O- + D-O-D <=> H-O-D + O-D-}$$

If the acid/base reaction were slow compared to diffusion, there would be little effect on the diffusion constant of hydroxide, and it would be similar to that of fluoride. However, because the main mechanism of hydroxide transport is by acid/base (i.e. the protons move and the oxygen atoms don't have to), you get a more than four-fold difference. This means acid/base reactions are on a time scale faster than diffusion.

This is also reflected in the high molar conductivities, as stated in the comment by Poutnik.

So as stated in other answers, the exchange will be fast compared to the time required to mix the two liquids.

The analogous mechanism would work for $\ce{H3O+}$ but it is harder to compare to other ions, so I started with hydroxide. You can imagine that the rates are heavily dependent on temperature and pH. However, even at neutral pH the rate is fast.

Does some "easy" way exist to estimate the reaction speed in such systems, or it has to be determined experimentally?

Here is a paper where they used computational chemistry to look at this question: http://omh.umeche.maine.edu/pdfs/JChemPhys_135_124505.01pdf.pdf. I would not consider this "easy", though.

If you compare the diffusion coefficient of hydroxide ($\pu{5.270e9 m^2/s}$) to that of fluoride ($\pu{1.460e9 m^2/s}$), you might be surprised to see such a difference despite their comparable size. The mechanism for the higher diffusion rate are acid/base reactions with neighboring waters. Instead of swapping positions, a water donates a proton to hydroxide, turning into hydroxide.

$$\ce{H-O- + H-O-H <=> H-O-H + O-H-}$$

With one deuterated water, it leads to formation of the desired product:

$$\ce{H-O- + D-O-D <=> H-O-D + O-D-}$$

If the acid/base reaction were slow compared to diffusion, there would be little effect on the diffusion constant of hydroxide, and it would be similar to that of fluoride. However, because the main mechanism of hydroxide transport is by acid/base (i.e. the protons move and the oxygen atoms don't have to), you get a more than four-fold difference. This means acid/base reactions are on a time scale faster than diffusion.

This is also reflected in the high molar conductivities, as stated in the comment by Poutnik.

So as stated in other answers, the exchange will be fast compared to the time required to mix the two liquids.

The analogous mechanism would work for $\ce{H3O+}$ but it is harder to compare to other ions, so I started with hydroxide. You can imagine that the rates are heavily dependent on pH. However, even at neutral pH the rate is fast.

If you compare the diffusion coefficient of hydroxide ($\pu{5.270e9 m^2/s}$) to that of fluoride ($\pu{1.460e9 m^2/s}$), you might be surprised to see such a difference despite their comparable size. The mechanism for the higher diffusion rate are acid/base reactions with neighboring waters. Instead of swapping positions, a water donates a proton to hydroxide, turning into hydroxide.

$$\ce{H-O- + H-O-H <=> H-O-H + O-H-}$$

With one deuterated water, it leads to formation of the desired product:

$$\ce{H-O- + D-O-D <=> H-O-D + O-D-}$$

If the acid/base reaction were slow compared to diffusion, there would be little effect on the diffusion constant of hydroxide, and it would be similar to that of fluoride. However, because the main mechanism of hydroxide transport is by acid/base (i.e. the protons move and the oxygen atoms don't have to), you get a more than four-fold difference. This means acid/base reactions are on a time scale faster than diffusion.

This is also reflected in the high molar conductivities, as stated in the comment by Poutnik.

So as stated in other answers, the exchange will be fast compared to the time required to mix the two liquids.

The analogous mechanism would work for $\ce{H3O+}$ but it is harder to compare to other ions, so I started with hydroxide. You can imagine that the rates are heavily dependent on temperature and pH. However, even at neutral pH the rate is fast.

Does some "easy" way exist to estimate the reaction speed in such systems, or it has to be determined experimentally?

Here is a paper where they used computational chemistry to look at this question: http://omh.umeche.maine.edu/pdfs/JChemPhys_135_124505.01pdf.pdf. I would not consider this "easy", though.