# What is the molarity of a solution of nitric acid if 0.216 of barium hydroxide is required to neutralize 20.00mL of nitric acid?

Nov 9, 2015

I assume that it is $0.216$ $g$ of barium hydroxide, and $20.00$ $m L$ nitric acid.

#### Explanation:

$0.216$ $g$ $B a {\left(O H\right)}_{2}$ $=$ $\frac{0.216 \cdot g}{171.34 \cdot g \cdot m o {l}^{-} 1}$ $=$ $1.26 \times {10}^{- 3}$ $m o l$ $B a {\left(O H\right)}_{2}$.

Now, of course I need a chemical equation:

$B a {\left(O H\right)}_{2} \left(a q\right) + 2 H N {O}_{3} \rightarrow B a {\left(N {O}_{3}\right)}_{2} \left(a q\right) + 2 {H}_{2} O \left(l\right)$.

Given the stoichiometry of the equation above (what do I mean here?), in $20.00$ $m L$ solution, there were $2.52 \times {10}^{- 3}$ moles of nitric acid present.

So concentration of the $H N {O}_{3} \left(a q\right)$ $=$ $\frac{n}{V}$ $=$ $\frac{2.52 \times {10}^{- 3} m o l}{20.00 \times {10}^{- 3} L}$ $\cong$ $0.1 \cdot m o l \cdot {L}^{-} 1$. You'll have to work out the actual figure; my calculator is never handy.