# A 15.62*mL volume of monoprotic acid reached a stoichiometric endpoint with 16.49*mL of sodium hydroxide whose concentration was 1.175*mol*L^-1. What is the concentration of the acid?

Sep 9, 2017

$\left[H X\right] = 1.24 \cdot m o l \cdot {L}^{-} 1$

#### Explanation:

We need (i) a stoichiometric equation,

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

And thus there is 1:1 equivalence between acid and base.

By definition, $\text{concentration"="moles of solute"/"volume of solution}$...

And thus the PRODUCT, $\text{volume"xx"concentration"="moles}$

And so we need (ii) molar quantities of acid and base.....

$\text{Moles of NaOH} \equiv 16.49 \times {10}^{-} 3 \cdot L \times 1.175 \cdot m o l \cdot {L}^{-} 1 = 0.01938 \cdot m o l$

But given the 1:1 equivalence between acid and base, CLEARLY this molar quantity was contained in the original $15.62 \cdot m L$ volume....

And so (finally)......

["Base"]=(0.01938*mol)/(15.62xx10^-3*L)=??*mol*L^-1

And note that doing the calculation this way, i.e. dimensionally with units included, gives me the appropriate units if we set up the problem correctly.