Question

A 0.753 g sample of a monoprotic acid is dissolved in water and titrated with 0.250 M NaOH. What is the molar mass of the acid if 21.5 mL of the NaOH solution is required to neutralize the sample?

Answer 1

`MM_(acid)=140.1g/mol`

Step-by-step explanation:

Hello,

In this case, since the acid is monoprotic, we can notice a 1:1 molar ratio between, therefore, for the titration at the equivalence point, we have:

`n_(acid)=n_(base) \\\\V_(acid)M_(acid)=V_(base)M_(base)\\\\n_(acid)=V_(base)M_(base)`

Thus, solving for the moles of the acid, we obtain:

`n_(acid)=0.0215L*0.250(mol)/(L)=5.375x10^(-3)mol`

Then, by using the mass of the acid, we compute its molar mass:

`MM_(acid)=(0.753g)/(5.375x10^(-5)mol) \\\\MM_(acid)=140.1g/mol`

Regards.