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

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 = $0.256 M \times 0.03378 L$

                                = $8.64 \times 10^{-3}$

                                = 0.00864 mol

     Moles of HA = 0.00864

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

      Molecular weight = $\frac{1.056 g}{0.00864 mol}$

                                    = 122.22 g/mol

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