A \(10.00-\mathrm{g}\) sample of the ionic compound NaA, where \(\mathrm{A}^{-}\) is the anion of a weak acid, was dissolved in enough water to make \(100.0 \mathrm{~mL}\) of solution and was then titrated with \(0.100 M\) HCl. After \(500.0 \mathrm{~mL}\) HCl was added, the \(\mathrm{pH}\) was \(5.00\). The experimenter found that \(1.00 \mathrm{~L}\) of \(0.100 \mathrm{M} \mathrm{HCl}\) was required to reach the stoichiometric point of the titration. a. What is the molar mass of \(\mathrm{NaA}\) ? b. Calculate the \(\mathrm{pH}\) of the solution at the stoichiometric point of the titration.

\ To find the molar mass of NaA, we first need to determine the moles of HCl at the stoichiometric point. We are given the volume (1.00L) and concentration (0.100M) of HCl at the stoichiometric point: Moles of HCl = volume × concentration Moles of HCl = 1.00 L × 0.100 M = 0.100 mol

\ At the stoichiometric point, the number of moles of A¯ will be equal to the number of moles of H⁺ from HCl. So, the moles of A¯ in the solution are 0.100 mol.

\ The 10.00g sample of the NaA compound was dissolved in 100.0 mL of water before performing the titration. At this point, the concentration of A¯ is equal to the moles of A¯ divided by the volume of the solution in liters: Concentration of A¯ = moles of A¯ / volume of solution Concentration of A¯ = 0.100 mol / 0.100 L = 1.00 M

\ We now have the concentration of A¯ in the solution and the mass of the compound. We can use these values to calculate the molar mass of NaA: Molar mass of NaA = mass of NaA / moles of A¯ Molar mass of NaA = 10.00 g / 0.100 mol = 100.00 g/mol The answer for part a is: The molar mass of NaA is 100.00 g/mol.

\ At the stoichiometric point, the moles of A¯ and HA are equal. Since we already know the moles of A¯ at the stoichiometric point (0.100 mol), we can use the volume of the solution at that point to calculate the concentration of HA: Concentration of HA = moles of HA / volume of solution at stoichiometric point Concentration of HA = 0.100 mol / (0.100 L + 1.00 L) = 0.100 mol/1.10 L = 0.0909 M

\ At the stoichiometric point, the solution contains only the weak acid HA. We can use its concentration (0.0909 M) and the given pH (5.00) at a certain point of the titration (after addition of 500.0 mL of HCl) to determine the Ka of the weak acid: \[Ka = 10^{-pKa}\] \[pKa = pH - \log{(\frac{[HA]}{[A^-]})}\] \[pKa = 5.00 - \log{(\frac{0.0909 M}{0.100 M})}\] \[pKa \approx 4.95\] \[Ka \approx 1.12 \times 10^{-5}\] Now we can use the Ka value and the concentration of HA to calculate the H⁺ concentration at the stoichiometric point: \[Ka = \frac{[H^+][A^-]}{[HA]}\] \[1.12 \times 10^{-5} = \frac{[H^+]^2}{0.0909 M}\] \[H^+ = \sqrt{1.12 \times 10^{-5} \times 0.0909 M}\] \[H^+ \approx 1.02 \times 10^{-3}\] Finally, we can calculate the pH at the stoichiometric point using the H⁺ concentration: \[pH = -\log{[H^+]}\] \[pH \approx -\log{(1.02 \times 10^{-3})}\] \[pH \approx 2.99\] The answer for part b is: The pH of the solution at the stoichiometric point is 2.99.