A sample of \(0.1687 \mathrm{~g}\) of an unknown monoprotic acid was dissolved in \(25.0 \mathrm{~mL}\) of water and titrated with \(0.1150 \mathrm{M} \mathrm{NaOH}\). The acid required \(15.5 \mathrm{~mL}\) of base to reach the equivalence point. (a) What is the molecular weight of the acid? (b) After \(7.25 \mathrm{~mL}\) of base had been added in the titration, the \(\mathrm{pH}\) was found to be 2.85 . What is the \(K_{a}\) for the unknown acid?

At the equivalence point, the number of moles of NaOH is equal to the number of moles of the unknown acid. We can find the moles of NaOH using the volume of NaOH titrated and its molarity. Moles of NaOH = Molarity × Volume Moles of NaOH = 0.1150 M × 15.5 mL × (1 L / 1000 mL) Moles of NaOH = 0.0017825 mol

Since the moles of NaOH and the acid are equal at the equivalence point, the moles of the unknown acid are also 0.0017825 mol.

We can use the mass and moles of the acid to calculate the molecular weight. Molecular weight = mass / moles Molecular weight = 0.1687 g / 0.0017825 mol Molecular weight ≈ 94.6 g/mol So, the molecular weight of the acid is approximately 94.6 g/mol.

Now we can calculate the molarity of the acid using the volume of water it is dissolved in: Molarity of the acid = moles / volume Molarity of the acid = 0.0017825 mol / 0.025 L Molarity of the acid = 0.0713 M

We'll use the same molarity formula to calculate the moles of NaOH present after 7.25 mL of base has been added: Moles of NaOH = Molarity × Volume Moles of NaOH = 0.1150 M × 7.25 mL × (1 L / 1000 mL) Moles of NaOH = 0.00083375 mol

To calculate the moles of the acid after 7.25 mL of base has been added, we subtract the moles of NaOH from the initial moles of the acid: Moles of the acid = Initial moles - moles of NaOH Moles of the acid = 0.0017825 mol - 0.00083375 mol Moles of the acid = 0.00094875 mol

We have the pH and the moles of the acid and base after 7.25 mL of base has been added. We can use the Henderson-Hasselbalch equation to find the Ka: \[pH = pKa + \log \frac{[A^-]}{[HA]}\] Rearrange the equation to find pKa: \[pKa = pH - \log \frac{[A^-]}{[HA]}\] Since we know the pH, we need to find the concentrations of [A^-] and [HA]. The total volume of the solution at this point is the sum of the initial volume of the acid (25 mL) and the volume of NaOH added (7.25 mL): Total volume = 25 mL + 7.25 mL = 32.25 mL = 0.03225 L [A^-] = moles of NaOH / total volume [A^-] = 0.00083375 mol / 0.03225 L [A^-] = 0.02585 M [HA] = moles of the acid / total volume [HA] = 0.00094875 mol / 0.03225 L [HA] = 0.02942 M Now we can find the pKa: \[pKa = 2.85 - \log \frac{0.02585}{0.02942}\] \[pKa ≈ 2.71\] Finally, to find the Ka, we use: \[Ka = 10^{-pKa}\] \[Ka ≈ 1.95 × 10^{-3}\] So, the Ka of the unknown acid is approximately 1.95 × 10^{-3}.