# What volume of 0.500 M NaOH is needed to neutralize 45.0 mL of 0.400 M HCl?

May 3, 2016

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

#### Explanation:

Clearly there is a 1:1 equivalence, and as a first step we calculate the number of moles of hydrochloric acid:

$45.0 \times {10}^{-} 3 \cdot L \times 0.400 \cdot m o l \cdot {L}^{-} 1$ $=$ $1.80 \times {10}^{-} 2 \cdot m o l \text{ hydrochloric acid}$.

We find an equivalent molar quantity of sodium hydroxide:

$\frac{1.80 \times {10}^{-} 2 \cdot m o l}{0.500 \cdot m o l \cdot {L}^{-} 1} \times {10}^{3} \cdot m L \cdot {L}^{-} 1$ $=$ $36.0 \cdot m L$.

This volume is reasonable, in that it is LESS than the volume of the less concentrated acid.