Question

What volume, in milliliters, of 0.530 M Ca(OH)2 is needed to completely neutralize 276 mL of a 0.220 M HI solution?

Answer 1

This is again a question that is completely made-up...which I think is a no-no....

Explanation:

I don't think you can prepare a `Ca(OH)_2(aq)` solution whose concentration is `0.530*mol*L^-1`. For calcium hydroxide, `K_"sp"=5.5xx10^-6`, and this gives a max. solubility of approx. `0.02*mol*L^-1`. And whoever set the question needs to get his/her act together...

I propose that the solution is `0.530*mol*L^-1` in `NaOH(aq)`...the which is chemically reasonable and reacts according to the stoichiometric equation...

`NaOH(aq) + HI(aq) rarr NaI(aq) + H_2O(l)`

`"Moles of HI(aq)"=276*mLxx10^-3*L*mL^-1xx0.220*mol*L^-1=0.06072*mol`.

And we require ONE equiv of sodium hydroxide solution....as is manifestly clear from the given stoichiometry.

And so `"volume of NaOH(aq)"=(0.06072*mol)/(0.530*mol*L^-1)xx10^3*mL*L^-1=114.6*mL`

The hypothetical calcium hydroxide solution would require DOUBLE this volume.....

Now clearly I answered the question I wanted to, and not the question you asked, but the question you asked was chemically unreasonable, and should not have been proposed...tell your teacher to lift his game....