# What volume of 0.475 M hydrochloric acid solution reacts with 36.4 mL of 0.350 M sodium hydroxide solution?

May 6, 2017

We interrogate the stoichiometric reaction...........and get a volume of approx. $27 \cdot m L$ of the given acid.

#### Explanation:

We interrogate the stoichiometric reaction...........

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

The reaction has 1:1 stoichiometry, and thus moles of acid are equivalent to moles of base.

Now $\text{concentration"="moles of solute"/"volume of solution}$;

And thus $\text{moles of solute"="concentration"xx"volume of solution}$.

And thus.....................................

${n}_{\text{NaOH}} = 36.4 \times {10}^{-} 3 \cdot L \times 0.350 \cdot m o l \cdot {L}^{-} 1 = 1.27 \times {10}^{-} 2 \cdot m o l$

And given the molar quantity, we may this work out the required volume of $H C l$:

$\text{Volume"_"HCl} = \frac{1.27 \times {10}^{-} 2 \cdot m o l}{0.475 \cdot m o l \cdot {L}^{-} 1} \times 1000 \cdot m L \cdot {L}^{-} 1 =$

$= 26.8 \cdot m L$........