# If it takes 25 mL of 0.05 M HCl to neutralize 345 mL of NaOH solution, what is the concentration of the NaOH solution?

Jun 27, 2016

$\left[N a O H\right] \cong 4 \times {10}^{-} 3 \cdot m o l \cdot {L}^{-} 1$

#### Explanation:

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

There is 1:1 equivalence between acid and base.

Thus $\left[N a O H\right]$ $=$ $\frac{25 \times {10}^{-} 3 L \times 0.05 \cdot m o l \cdot {L}^{-} 1}{345 \times {10}^{-} 3 L}$

$=$ ??*mol*L^-1