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#?

1 Answer
anor277
Oct 22, 2017

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.

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