If 25.0 mL of 0.100 M HClHCl is titrated with 0.150 M Ba(OH)_2Ba(OH)2. what volume of barium hydroxide is required to neutralize the acid?

1 Answer
anor277
Jan 19, 2017

A little less than half the volume of the starting hydrochloric acid.

Explanation:

We need (i) a stoichiometric equation:

Ba(OH)_2(aq) + 2HCl(aq) rarr BaCl_2(aq) + 2H_2OBa(OH)2(aq)+2HCl(aq)→BaCl2(aq)+2H2O,

Which shows us that 2 equiv acid react with the 1 equiv of barium hydroxide.

And (ii) we need the equivalent quantities of acid and base, observing the stoichiometry.

"Moles of hydrochloric acid"Moles of hydrochloric acid == 25.00xx10^-3*Lxx0.100*mol*L^-1=2.50xx10^-3*mol25.00×10−3⋅L×0.100⋅mol⋅L−1=2.50×10−3⋅mol

Given the stoichiometry, this molar quantity represents HALF of the molar quantity of barium hydroxide, which of course had an initial concentration of 0.150*mol*L^-10.150⋅mol⋅L−1. So we take the quotient,

1/2xx(2.50xx10^-3*mol)/(0.150*mol*L^-1)xx10^3*mL*L^-1~=8*mL12×2.50×10−3⋅mol0.150⋅mol⋅L−1×103⋅mL⋅L−1≅8⋅mL

I have not checked the solubility of the barium salt. Barium hydroxide has limited aqueous solubility. Presumably this question is consistent with experiment.

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