A \(10.00-g\) 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 mL of solution and was then titrated with 0.100 \(M\) HCl. After 500.0 mL HCl was added, the pH was \(5.00 .\) The experimenter found that 1.00 L of \(0.100 M\) HCl was required to reach the stoichiometric point of the titration. a. What is the molar mass of NaA? b. Calculate the \(p\) H of the solution at the stoichiometric point of the titration.

First, we'll find the moles of HCl added at the stoichiometric point and moles of NaA in the original solution. Given: Volume of HCl required to reach the stoichiometric point = 1.00 L Concentration of HCl = 0.100 M Moles of HCl = Concentration × Volume Moles of HCl = 0.100 moles/L × 1.00 L Recall that for a complete reaction, moles of NaA = moles of HCl.

Now that we have the moles of NaA, we can find its molar mass. Given: Mass of NaA = 10.00 g Moles of NaA = Moles of HCl = 0.100 moles Molar mass of NaA = (Mass of NaA)/(Moles of NaA) = (10.00 g)/(0.100 moles) = 100.00 g/mol Answer to part (a): The molar mass of NaA is 100.00 g/mol.

To calculate the pH at the stoichiometric point, we need to find the concentration of the A^(-) anion. The stoichiometric point reached means that all NaA has reacted with HCl, so the remaining solution will have A^(-): \(NaA + HCl \rightarrow NaCl + HA\) At this point, we can use the given pH value before the stoichiometric point was reached to find the Ka (Acid dissociation constant) of the weak acid HA. We must remember that the concentration of A^(-) is equal to the concentration of H+ at this point. Given: pH_after_500mL = 5.00 pKa = -log(Ka) From the given pH value, we get: [H+] = 10^(-5.00) = 1.0 × 10^(-5) M Following the formula, pKa = -log([H+] × [A^(-)]/[HA]) Adding 500 mL of HCl to 100 mL of NaA solution, the total volume will be 600 mL or 0.6 L. We calculate the concentration of A^(-), which is equal to [H+] at this point. [A^(-)] = (0.100 M × 0.500 L) / 0.6 L = 0.0833 M We can find the concentration of HA by considering that in the reaction, 1 mole of NaA is converted to 1 mole of HA. [HA] = (0.100 M × 0.100 L) / 0.6 L = 0.0167 M Now we can calculate pKa: 5.00 - pH_after_500mL = -log((1.0 × 10^(-5) × 0.0833)/0.0167) pKa = 4.83 Therefore, Ka = 10^(-4.83) At the stoichiometric point, the remaining solution composition is [A^(-)] = [HA] = 0.100 M (since they are produced in equimolar amounts, and NaCl does not affect the pH). We can now use the Ka expression to find [H+] at the stoichiometric point: Ka = [H+][A^(-)]/[HA] [H+] = Ka * [HA]/[A^(-)] = 10^(-4.83) * 0.100/0.100 = 10^(-4.83) M Now, we can find the pH at the stoichiometric point by taking the negative logarithm of the [H+] concentration: pH = -log10([H+]) = -log10(10^(-4.83)) = 4.83 Answer to part (b): The pH at the stoichiometric point of the titration is 4.83.