Question

An `80*mL` volume of `NaOH` of `1.00*mol*L^-1` concentration was stoichiometrically equivalent to a `70*mL` volume of `HCl`. What is `[HCl]`?

Answer 1

`[HCl]=1.14*mol*L^-1`, a tad more concentrated than is the base...

Explanation:

We write the stoichiometric equation that represents the reaction between hydrochloric acid, and sodium hydroxide....

`NaOH(aq) + HCl(aq) rarr NaCl(aq) + H_2O(l)`

And we do this to establish the 1:1 stoichiometry....and so then we use the definition of concentration....

`"Concentration"="Moles of solute (mol)"/"Volume of solution (L)"`

And so....

`"moles of NaOH(aq)"=80.0*mLxx10^-3*L*mL^-1xx1.00*mol*L^-1=0.0800*mol`

And there was an equivalent quantity of moles in a `70.0*mL` volume of `HCl(aq)`...

`[HCl(aq)]=(0.0800*mol)/(70.0*mLxx10^-3*L*mL^-1)=1.143*mol*L^-1`