Calculate the \(\mathrm{pH}\) at the halfway point and at the equivalence point for each of the following titrations. $$ \begin{array}{l}{\text { a. } 100.0 \mathrm{mL} \text { of } 0.10 \mathrm{M} \mathrm{HC}_{7} \mathrm{HsO}_{2}\left(K_{\mathrm{a}}=6.4 \times 10^{-5}\right) \text { titrated }} \\ {\text { by } 0.10 \mathrm{M} \mathrm{NaOH}}\end{array} $$ $$ \begin{array}{l}{\text { b. } 100.0 \mathrm{mL} \text { of } 0.10 M \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{2}\left(K_{\mathrm{b}}=5.6 \times 10^{-4}\right) \text { titrated }} \\ {\text { by } 0.20 \mathrm{M} \mathrm{HNO}_{3}}\end{array} $$ $$ 100.0 \mathrm{mL} \text { of } 0.50 \mathrm{M} \mathrm{HCl} \text { titrated by } 0.25 \mathrm{M} \mathrm{NaOH} $$

To calculate the initial moles of the weak acid, use the formula: moles = molarity × volume, which in this case is (0.10 mol/L)(0.100 L) = 0.01 moles.

At the halfway point, we have added half the moles of NaOH required to reach the equivalence point. Thus, the number of moles of NaOH is 0.005 moles. At the equivalence point, the moles of added NaOH are equal to the initial moles of the weak acid, which is 0.01 moles.

At the halfway point, we treat the solution as a buffer. This is because the number of moles of the weak acid equals the moles of its conjugate base. We can calculate the pH using the Henderson-Hasselbalch equation: pH = pKa + log ([A-]/[HA]). In this case, pKa is calculated as -log(Ka), which is -log(6.4 × 10^(-5)) = 4.19. Since [A-] = [HA] at the halfway point, the pH = pKa = 4.19.

At the equivalence point, we have the presence of the conjugate base C7H5O2^- (A^-) in the Solution. We will ignore any effects of H3O^+ contribution from water, as the effect is minimal. To determine the pH, we need to treat the solution as if it is a solution of a weak base. We use Kb for the dissociation of the weak base: Kb = Kw / Ka, where Kw = 1×10^(-14). Kb = (1×10^(-14))/(6.4×10^(-5)) = 1.56×10^(-10). Setting up an ICE table and solving for the hydroxide ion [OH^-], we get: [OH^-] = sqrt((1.56×10^(-10))(0.10 M)) = 1.25×10^(-6) M. To find the pH, first, we calculate pOH = -log[OH^-] = 5.90, and then we calculate pH by using the formula pH = 14 - pOH = 14 - 5.90 = 8.10. #For scenario b (weak base and strong acid):# Follow a similar process as in scenario a, but this time, consider the weak base and its conjugate acid. Calculate the values required for the Kb to obtain the pOH and convert it into pH at the halfway point and the equivalence point. #For scenario c (strong acid and strong base):#

To calculate the initial moles of the strong acid, use the formula: moles = molarity × volume, which in this case is (0.50 mol/L)(0.100 L) = 0.050 moles of HCl.

At the halfway point, we have added half the moles of NaOH required to reach the equivalence point. Thus, the number of moles of NaOH is 0.025 moles. At the equivalence point, the moles of added NaOH are equal to the initial moles of the strong acid, which is 0.050 moles.

At the halfway point, use the concept of a buffer solution, as the number of moles of the strong acid equals the moles of its conjugate base. First, we calculate [H3O^+] = (initial moles of HCl)/(total volume in liters) = 0.025 moles / (0.100 L + 0.100 L) = 0.125 mol/L. Then, we calculate pH = -log[H3O^+] = -log(0.125) = 0.90.

At the equivalence point, there is only water present, because strong acids and strong bases neutralize each other completely. So, the pH of the solution will be 7, as it is a neutral solution.