A \(10.00-\mathrm{g}\) sample of the ionic compound \(\mathrm{NaA},\) 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 \mathrm{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 NaA? b. Calculate the \(\mathrm{pH}\) of the solution at the stoichiometric point of the titration.

The volume of \(0.100 \mathrm{M}\) HCl required to reach the stoichiometric point is \(1.00 \mathrm{L}\). We can use the concentration and volume to find the moles of HCl: Moles of HCl = (Concentration of HCl) x (Volume of HCl) Moles of HCl = \(0.100 \mathrm{M} \times 1.00 \mathrm{L} = 0.100 \mathrm{mol}\)

At the stoichiometric point, the moles of HCl and A- are equal. Therefore, the moles of A- = moles of HCl = \(0.100 \mathrm{mol}\).

Since NaA dissociates into one Na+ and one A- ion, the moles of Na+ = moles of A- = \(0.100 \mathrm{mol}\).

To find the molar mass of NaA, divide the mass of the sample by the moles of Na+: Molar mass of NaA = \(\frac{10.00 \mathrm{g}}{0.100 \mathrm{mol}} = 100.00 \mathrm{g/mol}\)

At the stoichiometric point, the amounts of HCl and A− are equal, such that all the A− is converted into HA (the weak acid). Following the stoichiometric point, there is an excess of HCl in the solution. Since the \(\mathrm{pH}\) of the acidic solution depends on the \(\mathrm{pH}\) of the strong acid (HCl) in the solution, we can calculate the concentration of HCl at the stoichiometric point: Concentration of excess HCl = \(\frac{0.100 \mathrm{mol} \times (1.00 \mathrm{L} - 0.100 \mathrm{L})}{0.100 \mathrm{L}} = 0.0900 \mathrm{M}\) Now, we can find the \(\mathrm{pH}\) using the concentration of HCl: \(\mathrm{pH}\) = \(-\log (0.0900) \approx 1.05\) The \(\mathrm{pH}\) of the solution at the stoichiometric point of the titration is approximately 1.05. In summary: a. The molar mass of NaA is \(100.00 \mathrm{g/mol}\). b. The \(\mathrm{pH}\) of the solution at the stoichiometric point of the titration is approximately \(1.05\).