# A 60*mL volume of sodium hydroxide at 0.15*mol*L^-1 concentration is titrated with 0.500*mol*L^-1 HCl(aq). What volume of acid is required?

May 25, 2017

We use the relationship ${C}_{1} {V}_{1} = {C}_{2} {V}_{2}$, where ${C}_{i}$, and ${V}_{i}$ are the individual concentrations/volumes. But FIRST you need a stoichiometric equation. Eventually I gets under $20 \cdot m L$.

#### Explanation:

Here, there is certainly 1:1 stoichiometry between acid and base, i.e.:

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

And please note that the product ${C}_{1} {V}_{1} = \text{moles} \cdot {L}^{-} 1 \times L$, i.e. an ANSWER in moles as required...........

And so if ${C}_{1} {V}_{1} = {C}_{2} {V}_{2}$, then............

${V}_{2} = \frac{{C}_{1} {V}_{1}}{C} _ 2 = \frac{60 \cdot m L \times 0.15 \cdot m o l \cdot {L}^{-} 1}{0.500 \cdot m o l \cdot {L}^{-} 1} = 18 \cdot m L$