Question

Given that `[Ba(OH)_2]=0.163*mol*L^-1`, what volume of this solution is required to neutralize a `13.8*mL` volume of `HCl` whose concentration is `0.170*mol*L^-1`?

Answer 1

We use the relationship....`"Concentration"="Moles of solute"/"Volume of solution"`

Explanation:

And of course we need a stoichiometrically balanced equation....

`Ba(OH)_2(s) + 2HCl(aq) rarr BaCl_2(aq) + 2H_2O(l)`

Now barium hydroxide HAS limited solubility in water, certainly not to the extent of `0.163*mol*L^-1`; we will work out the MASS of barium hydroxide required for neutralization of the acid....

`"Moles of HCl"=13.8*mLxx10^-3*L*mL^-1xx0.170*mol*L^-1=2.35xx10^-3*mol`.

And this will neutralize HALF an equiv of `Ba(OH)_2(s)`....

i.e. `2.35xx10^-3*molxx1/2xx171.34*g*mol^-1=0.200*g`....

Please draw to your teachers attention, the impossibility of the reaction as written. Equations follow chemical reactions; chemical reactions do not follow equations.