# 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
Oct 22, 2017

We use the relationship....$\text{Concentration"="Moles of solute"/"Volume of solution}$

#### Explanation:

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

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

Now barium hydroxide HAS limited solubility in water, certainly not to the extent of $0.163 \cdot m o l \cdot {L}^{-} 1$; we will work out the MASS of barium hydroxide required for neutralization of the acid....

$\text{Moles of HCl} = 13.8 \cdot m L \times {10}^{-} 3 \cdot L \cdot m {L}^{-} 1 \times 0.170 \cdot m o l \cdot {L}^{-} 1 = 2.35 \times {10}^{-} 3 \cdot m o l$.

And this will neutralize HALF an equiv of $B a {\left(O H\right)}_{2} \left(s\right)$....

i.e. $2.35 \times {10}^{-} 3 \cdot m o l \times \frac{1}{2} \times 171.34 \cdot g \cdot m o {l}^{-} 1 = 0.200 \cdot g$....

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