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?

1 Answer
Sep 9, 2017

Answer:

#[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.