Question

Calculate the volume of 0.450 M Ba(OH)2 which will be needed to neutralize 46.00mL of 0.252 M HCl. What's the volume of the Ba(OH)2 solution?

Again I got an answer but I am not too sure on it. I got 0.0129L or 12.9mL. I would really appreciate a 2nd opinion if possible.

Answer 1

Well `"concentration"="moles of solute"/"volume of solution"`

Explanation:

And so by multiplication or division, GIVEN TWO of the THREE QUANTITIES, we can assess the third. But we need as a preliminary, a stoichiometrically balanced equation so as to inform our arithmetic, i.e. the stoichiometric equivalence... And so...

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

That is TWO EQUIV of `HCl` are required for reaction with ONE EQUIV of barium hydroxide.

`"Moles of HCl(aq)"=46.00*mLxx10^-3*L*mL^-1xx0.252*mol*L^-1=0.011592*mol`

...and THUS we need HALF an EQUIV of `Ba(OH)_2`...

And so...`(1/2xx0.011592*mol)/(0.450*mol*L^-1)xx1000*mL*L^-1=12.9*mL`...so I agree with you mate...

By the way, I think that barium hydroxide is NOT soluble to the extent of `0.450*mol*L^-1` in aqueous solution, and so your chemistry teacher needs to lift his/her game...