Question

In a titration, 16.02 mL of 0.100M NaOH was required to titrate 0.2011 g of an unknown acid, HX. What is the molar mass of the acid?

Answer 1

We assume the acid to be monoprotic.

Explanation:

Rxn: `HX(aq) + NaOH(aq) rarr H_2O(l) + NaX(aq)`

Therefore, moles of NaOH used `-=` moles of HX, the unknown acid.

Moles of NaOH:`16.02xx10^(-3)Lxx0.100*mol*L^(-1)=16.02xx10^(-4)* mol`.

Thus your sample mass represents `16.02xx10^(-4)*mol`.

The molar mass is simply the quotient, `(0.2011*g)/(16.02xx10^(-4)*mol)`, about `120*g*mol^(-1)?` At least I know it is right dimensionally, which is why I pfaff about with the units.