# A 2.0 mL sample of NaOH solution is exactly neutralized by 4.0 mL of 3.0 M HCl solution. What is the concentration of the NaOH solution?

##### 1 Answer
Jun 3, 2016

$\left[N a O H\right]$ $=$ $6.0 \cdot m o l \cdot {L}^{-} 1$

#### Explanation:

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

Clearly a 1:1 stoichiometry pertains. We know the concentration of the hydrochloric acid, and we use this to calculate (i) the number of moles of hydrochloric acid, and (ii) the concentration of the caustic soda.

$\text{Moles of hydrochloric acid}$ $=$ $4.0 \times {10}^{-} 3 L \times 3.0 \cdot m o l \cdot {L}^{-} 1$ $=$ $12 \times {10}^{-} 3 \cdot m o l$

$\text{Concentration of sodium hydroxide}$ $=$ $\text{Moles of NaOH"/"Initial volume of solution}$ $=$ $\frac{12 \times {10}^{-} 3 \cdot m o l}{2.0 \times {10}^{-} 3 \cdot L}$