Question

A solution of hydrochloric acid is prepared from a mass of `0.451*g` that is dissolved in `1.000*L` of water. For a `50.0*mL` aliquot of this solution, how much `0.0273*mol*L^-1` `HCl` titrant would be required for equivalence?

Answer 1

WE need approx. `23*mL` `NaOH(aq)` of the given concentration....

Explanation:

We always write a stoichiometric equation to inform our 'rithmetic...

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

There is thus 1:1 molar equivalence....

Now `[HCl]=((0.451*g)/(36.46*g*mol^-1))/(1.000*L)=0.0124*mol*L^-1`.

And we gots a `50*mL` volume of this stuff, i.e. a molar quantity of `50xx10^-3*Lxx0.0124*mol*L^-1=6.18xx10^-4*mol`

And since we titrate with `0.0273*mol*L^-1`, we take the quotient....

`(6.18xx10^-4*mol)/(0.0273*mol*L^-1)xx10^3*mL*L^-1=??mL`

Note that this is a good question with a sound practical basis, in that typically we would want to use a `20-30*mL` volume of titrant from a burette....