Question

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?

Answer 1

`[HX]=1.24*mol*L^-1`

Explanation:

We need (i) a stoichiometric equation,

`HX(aq) + NaOH(aq) rarr NaX(aq) + H_2O`

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

By definition, `"concentration"="moles of solute"/"volume of solution"`...

And thus the PRODUCT, `"volume"xx"concentration"="moles"`

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

`"Moles of NaOH"-=16.49xx10^-3*Lxx1.175*mol*L^-1=0.01938*mol`

But given the 1:1 equivalence between acid and base, CLEARLY this molar quantity was contained in the original `15.62*mL` 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.