# What volume of 0.365 N lithium hydroxide is needed to completely neutralize 27.68 mL of 0.559 N nitric acid?

Jan 2, 2016

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

#### Explanation:

Moles of nitric acid?

$27.68 \times {10}^{- 3} \cancel{L} \times 0.559 \cdot m o l \cdot \cancel{{L}^{- 1}}$ $=$ $1.54 \times {10}^{- 2}$ $m o l$.

So we need $\frac{1.54 \times {10}^{- 2} \cdot \cancel{m o l}}{0.365 \cdot \cancel{m o l} \cdot {L}^{-} 1}$ $=$ $4.24 \times {10}^{- 2}$ $L$

Or $4.24 \times {10}^{- 2}$ $\cancel{L}$ $\times$ $1000$ $m L$ $\cancel{{L}^{-} 1}$ $=$ ?? $m L$