Question

What volume of `0.373*mol*L^-1` nitric acid is required to react with a `17.5*mL` solution of `Ca(OH)_2` whose concentration is `0.177*mol*L^-1`?

Answer 1

We use the quotient `"concentration"="moles of solute"/"volume of solution"`

Explanation:

And as always with these sorts of questions we are obliged to write a stoichiometrically balanced equation that establishes mass equivalence....

`Ca(OH)_2(aq) + 2HNO_3(aq)`

`"Moles of calcium hydroxide"=17.5xx10^-3*Lxx0.177*mol*L^-1=3.10*mmol`.

And so we require 2 equiv of nitric acid.....

`V_"nitric acid"=(2xx3.10xx10^-3*mol)/(0.373*mol*L^-1)xx1000*mL*L^-1`

`=16.6*mL`

Just to add, I bet you cannot get a calcium hydroxide solution of this concentration, it is pretty insoluble stuff....you could point this out to your teacher to win brownie points. Of course we could add nitric acid to neutralize a given mass of the hydroxide.....which could be slurried up in solution....