Question

To determine the molar mass of an unknown organic acid, HA, a 1.056 g sample is titrated with standardized NaOH. Calculate the molar mass of HA assuming the acid reacts with 33.78 mL of 0.256 M NaOH. HA is monoprotic.

Answer 1

Step-by-step explanation:

Equation for the given reaction is as follows.

`HA(aq) + NaOH(aq) \rightarrow NaA(aq) + H_(2)O(l)`

Therefore, moles of NaOH and HA are calculated as follows.

Moles of NaOH =
`0.256 M * 0.03378 L`

=
`8.64 * 10^(-3)`

= 0.00864 mol

Moles of HA = 0.00864

Also, moles =
`\frac{\text{weight}}{\text{molecular weight}}`

Molecular weight =
`(1.056 g)/(0.00864 mol)`

= 122.22 g/mol

Thus, we can conclude that molar mass of given unknown organic acid is 122.22 g/mol.