# If 25.0 mL of 0.100 M HCl is titrated with 0.150 M Ba(OH)_2. what volume of barium hydroxide is required to neutralize the acid?

Jan 19, 2017

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

#### Explanation:

We need (i) a stoichiometric equation:

$B a {\left(O H\right)}_{2} \left(a q\right) + 2 H C l \left(a q\right) \rightarrow B a C {l}_{2} \left(a q\right) + 2 {H}_{2} O$,

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.

$\text{Moles of hydrochloric acid}$ $=$ $25.00 \times {10}^{-} 3 \cdot L \times 0.100 \cdot m o l \cdot {L}^{-} 1 = 2.50 \times {10}^{-} 3 \cdot m o l$

Given the stoichiometry, this molar quantity represents HALF of the molar quantity of barium hydroxide, which of course had an initial concentration of $0.150 \cdot m o l \cdot {L}^{-} 1$. So we take the quotient,

$\frac{1}{2} \times \frac{2.50 \times {10}^{-} 3 \cdot m o l}{0.150 \cdot m o l \cdot {L}^{-} 1} \times {10}^{3} \cdot m L \cdot {L}^{-} 1 \cong 8 \cdot m L$

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