# How many milliliters of 0.20N sodium hydroxide must be added to 75 mL of 0.050 N hydrochloric acid to make a neutral solution?

Apr 10, 2017

Approx. $19 \cdot m L$

#### Explanation:

We need (i) a stoichiometric equation in order to inform our calculations. Now with one equiv of simple acids and bases, this is trivial:

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

$\text{Moles of HCl} = 75 \times {10}^{-} 3 \cancel{L} \times 0.050 \cdot m o l \cdot \cancel{{L}^{-} 1}$

$3.75 \times {10}^{-} 3 \cdot m o l$ with respect to $H C l$.

And thus we need an equiv of $N a O H$, and we have $0.20 \cdot m o l \cdot {L}^{-} 1$ available, so..........we divide the molar quantity of $H C l$ by the $N a O H$ concentration to get the req'd volume

$\frac{3.75 \times {10}^{-} 3 \cdot m o l}{0.20 \cdot m o l \cdot {L}^{-} 1} = 0.01875 \cdot L = 18.8 \cdot m L$