# How many mL of a 3.2M NaOH solution is required to neutralize 74mL of a 12.4M HCl solution?

May 5, 2016

Approx. 300 mL

#### Explanation:

For all these sorts of problems we require a balanced chemical equation:

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

Such a scheme establishes that there is a 1:1 equivalence between moles acid and moles of base.

$\text{Moles of hydrochloric acid}$ $=$ $74 \times {10}^{-} 3 L \times 12.4 \cdot m o l \cdot {L}^{-} 1 = 0.895 \cdot m o l$.

Thus we require a quantity of $0.895 \cdot m o l \cdot \text{NaOH}$:

$\frac{0.895 \cdot \cancel{m o l}}{3.2 \cdot \cancel{m o l} \cdot {L}^{-} 1}$ $=$ $0.28 \cdot L \text{ sodium hydroxide solution}$

This quantity is reasonable as the the acid is 4 times as concentrated as the base.