How many mL of 0.200 M HCI are needed to neutralize 20.0 mL of 0.150 M Ba(OH)_2?

May 13, 2017

Approx. $\equiv 60 \cdot m L$

Explanation:

We use the relationship: $\text{Moles of stuff"/"Volume of solution"="Molarity}$.

There are $2 \times 20 \times {10}^{-} 3 \cdot L \times 0.150 \cdot m o l \cdot {L}^{-} 1 = 6.00 \times {10}^{-} 3 \cdot m o l$ $B a {\left(O H\right)}_{2}$, and this is neutralized according to the following equation:

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

Given the $\text{STOICHIOMETRY}$, we need 2 equiv of hydrochloric acid, and given $0.200 \cdot m o l \cdot {L}^{-} 1$ $H C l \left(a q\right)$ we need a volume of..........

$\frac{6.00 \times {10}^{-} 3 \cdot \cancel{m o l} \times 2}{0.200 \cdot \cancel{m o l} \cdot {L}^{-} 1} = 0.0600 \cdot L$

$\equiv 60 \cdot m L$