What volume of 0.20M Ca(OH)2 will neutralize 45.0 mL of a 1M solution of HClO3?

Well, the `"Ca"("OH")_2` will not all dissolve... so your question is problematic.

If you add `"HClO"_3` into `"Ca"("OH")_2` while mixing THOROUGHLY so that extra solid dissolves, then this problem may be more possible by Le Chatelier's principle (basic salts dissolve better in acidic solution), but if you add `"0.20 M"` `"Ca"("OH")_2` into `"HClO"_3`, that requires that the base FULLY dissolved first...

And so if you grab such a SATURATED solution, you only get `"0.011 M"`, and extra solid sinks to the bottom of the container you sourced it from.

The `K_(sp)` of `"Ca"("OH")_2` is `5.5 xx 10^(-6)`, whose mass action expression is

`5.5 xx 10^(-6) = ["Ca"^(2+)]["OH"^(-)]^2 = s(2s)^2 = 4s^3`

So, the solubility `s` is:

`s = ((5.5xx10^(-6))/4)^(1//3) = "0.011 M"`

Thus, the maximum concentration you can have is `"0.011 M"`, and the rest is well, a SOLID. The reaction is:

`"Ca"("OH")_2(aq) + 2"HClO"_3(aq) -> 2"H"_2"O"(l) + "Ca"("ClO"_3)_2(aq)`

In that volume of `"HClO"_3` you have:

`"1 M" cdot "0.045 L" = "0.045 mols HClO"_3 = "0.045 mols H"^(+)`

which react with

`0.045 cancel("mols OH"^(-)) xx ("1 mol Ca"("OH")_2)/cancel("2 mols OH"^(-)) = "0.0225 mols Ca"("OH")_2`

Therefore, in `"0.011 M"` of it, you would need

`"1 L"/(0.011 cancel"mol") xx 0.0225 cancel"mols" = color(blue)(ul"2.05 L")`

It sounds absolutely absurd, but `"Ca"("OH")_2` is NOT that strong of a strong base...

And in this `"2.05 L"`, there would have been `"28.71 g"` of `"Ca"("OH")_2(s)` that did NOT dissolve, that then falls back into the bottom of the container you sourced it from.