A sample of \(0.2140 \mathrm{~g}\) of an unknown monoprotic acid was dissolved in \(25.0 \mathrm{~mL}\) of water and titrated with \(0.0950 \mathrm{M} \mathrm{NaOH}\). The acid required \(27.4 \mathrm{~mL}\) of base to reach the equivalence point. (a) What is the molar mass of the acid? (b) After \(15.0 \mathrm{~mL}\) of base had been added in the titration, the \(\mathrm{pH}\) was found to be 6.50 . What is the \(K_{a}\) for the unknown acid?

To find the moles of NaOH, we will use the concentration and volume. Given the concentration of NaOH is \(0.0950\,M\) and the volume of NaOH required to reach the equivalence point is \(27.4\,mL\), we can calculate the moles of NaOH as follows: Moles of NaOH = Concentration × Volume Moles of NaOH = \(0.0950\,M\) × \(27.4\,mL\) × \(\frac{1L}{1000\, mL}\) Moles of NaOH = \(0.002603\,mol\)

Since it's a monoprotic acid, at the equivalence point, moles of acid (HA) = moles of NaOH. Thus, the moles of acid are also \(0.002603\,mol\).

We have the mass of the acid sample \(0.2140\,g\) and the moles of acid \(0.002603\,mol\). Using these, we can calculate the molar mass of the acid as follows: Molar mass = \(\frac{Mass\,of\,sample}{Moles\,of\,acid}\) Molar mass = \(\frac{0.2140\,g}{0.002603\,mol}\) Molar mass = \(82.21\,g/mol\) (approximately) So, the molar mass of the acid is about \(82.21\,g/mol\).

Before 15 mL of NaOH was added, the moles of HA were \(0.002603\,mol\). At 15 mL of NaOH, we can calculate the moles of NaOH used: Moles of NaOH at 15 mL = \(0.0950\,M\) × \(15\,mL\) × \(\frac{1\,L}{1000\,mL}\) = \(0.001425\,mol\) At this point, \(0.001425\,mol\) of acid (HA) has been neutralized by NaOH, leaving the remaining moles of acid: Moles of remaining HA = \(0.002603\,mol\) - \(0.001425\,mol\) = \(0.001178\,mol\) Now let's calculate the concentration of unreacted acid in the total volume of \(25\,mL\) (initial sample) + \(15\,mL\) (added NaOH) = \(40\,mL\): Concentration of remaining HA = \(\frac{0.001178\,mol}{40\,mL}\) × \(\frac{1\,L}{1000\,mL}\) ≈ \(0.02945\,M\)

With a given pH of 6.50, we can first calculate the H3O+ concentration: \[H_3O^+ = 10^{-\mathrm{pH}} = 10^{-6.50} = 3.16 \times 10^{-7}\, M\] Recall that the reaction for a weak monoprotic acid is: \(HA \rightleftharpoons H^+ + A^-\) At this stage in the titration with some added NaOH, we can assume that all the \(H^+\) ions come from the dissociation of HA, so the concentration of \(H^+\) and \(A^-\) is equal (\(3.16 \times 10^{-7}\,M\)). Using an ICE (Initial, Change, Equilibrium) table for the reaction, we get: Initial: [HA] = \(0.02945\,M\), [H+] = 0, [A-] = 0 Change: [HA] = -x, [H+] = +x, [A-] = +x Equilibrium: [HA] = \(0.02945-x\,M\), [H+] = x, [A-] = x Given: [H+] = [A-] = \(3.16 \times 10^{-7}\,M\), we have \(x = 3.16 \times 10^{-7}\,M\) Substituting the equilibrium values into the Ka expression: \[K_a = \frac{[\mathrm{H}^+][\mathrm{A}^-]}{[\mathrm{HA}]} = \frac{(3.16 \times 10^{-7})^2}{0.02945 - 3.16 \times 10^{-7}}\] Finally, calculate the value of Ka: "Ka ≈ \(3.39 \times 10^{-10}\)" The Ka of the unknown acid is approximately \(3.39 \times 10^{-10}\).