Question

What volume, in milliliters, of 0.54 M `Ca(OH)_2` is needed to completely neutralize 241.4 mL of a .26 M `HI` solution?

Answer 1

`Ca(OH)_2(aq) +2HI(aq)rarr CaI_2(aq) +2H_2O(l)`

Explanation:

`"Moles of HI"` `=` `241.4xx10^-3*Lxx0.26*mol*L^-1` `=` `0.063*mol*HI`

`0.0313*mol` `Ca(OH)_2` are required for equivalence. Now of course I could use this molar quantity to calulate the volume of calcium hydroxide required. However, there is one problem.

Calcium hydroxide is very sparingly soluble. I don't think you could make a `0.5*mol*L^-1` solution of this beast (I might be wrong, but I am not going to go thru the calculation). This question should have been proposed with `NaOH` solution