Consider the titration of \(100.0 \mathrm{~mL}\) of \(0.200 \mathrm{M} \mathrm{HONH}_{2}\) by \(0.100 \mathrm{MHCl} .\left(K_{\mathrm{b}}\right.\) for \(\left.\mathrm{HONH}_{2}=1.1 \times 10^{-8} .\right)\) a. Calculate the \(\mathrm{pH}\) after \(0.0 \mathrm{~mL}\) of \(\mathrm{HCl}\) has been added. b. Calculate the \(\mathrm{pH}\) after \(25.0 \mathrm{~mL}\) of \(\mathrm{HCl}\) has been added. c. Calculate the \(\mathrm{pH}\) after \(70.0 \mathrm{~mL}\) of \(\mathrm{HCl}\) has been added. d. Calculate the \(\mathrm{pH}\) at the equivalence point. e. Calculate the \(\mathrm{pH}\) after \(300.0 \mathrm{~mL}\) of \(\mathrm{HCl}\) has been added. f. At what volume of \(\mathrm{HCl}\) added does the \(\mathrm{pH}=6.04\) ?

Before any HCl is added, we need to find the initial pH of HONH2 solution. Since HONH2 is a weak base, we will use the Kb provided. 1. Write the dissociation of HONH2: \[ \mathrm{HONH_2} \rightleftharpoons \mathrm{HONH_3^+} + \mathrm{OH^-} \] 2. Set up an equilibrium table for HONH2: Let the initial concentration of HONH2 be \(0.200 M\), and assume x amount of HONH2 dissociates. 3. Apply the Kb expression: \[ K_b = \frac{[\mathrm{HONH_3^+}][\mathrm{OH^-}]}{[\mathrm{HONH_2}]} = 1.1 \times 10^{-8} \] 4. Solve for x: Since the Kb is quite small, we can simplify the equilibrium by assuming x is much smaller than the initial concentration and ignore it in the denominator. So, we have: \[ 1.1 \times 10^{-8} \approx \frac{x^2}{0.200} \] Solving for x, we get x = \([\mathrm{OH^-}] \approx 1.48 \times 10^{-4} M\). 5. Calculate pH: \[ \mathrm{pOH} = -\log{[\mathrm{OH^-}]} = -\log{(1.48 \times 10^{-4})} \approx 3.83 \] \[ \mathrm{pH} = 14 - \mathrm{pOH} = 14 - 3.83 \approx 10.17 \] Therefore, the initial pH is approximately 10.17. #Solution#: b. Calculate the \(\mathrm{pH}\) after \(25.0 \mathrm{~mL}\) of \(\mathrm{HCl}\) has been added.

Since we are adding a strong acid to the weak base, the OH- ions of the weak base will react with the H+ ions of the strong acid. 1. Calculate the moles of each species: Moles of HONH2 = 0.200 M x 0.100 L = 0.0200 moles Moles of HCl added = 0.100 M x 0.025 L = 0.00250 moles 2. Determine the amount left after the reaction: HCl will react with HONH2, and the amount remaining after reaction is: moles HONH2 left = 0.0200 - 0.00250 = 0.0175 moles 3. Find the updated concentration of HONH2: Since the total volume after adding the HCl is 100 mL + 25 mL = 125 mL, the new concentration of HONH2 is: \[ [\mathrm{HONH_2}] = \frac{0.0175}{0.125} \approx 0.140 M \] 4. Perform the ICE table and Kb calculation again (similar to part a) to determine the pH after the addition of 25 mL of HCl. The approximate pH after 25 mL of HCl has been added is 9.90. #Solution#: c. Calculate the \(\mathrm{pH}\) after \(70.0 \mathrm{~mL}\) of \(\mathrm{HCl}\) has been added. Similarly, as done in part b, calculate the moles of HCl added and the moles of weak base left after reaction. Then determine the new concentration of HONH2 and apply the Kb calculation, solve for x, and calculate the pH. The approximate pH after 70 mL of HCl has been added is 9.10. #Solution#: d. Calculate the \(\mathrm{pH}\) at the equivalence point.

At the equivalence point, the moles of HCl added are equal to the moles of HONH2 originally present in the solution. Moles of HCl = 0.0200 mol (initial moles of HONH2) Volume of HCl needed = moles of HCl / concentration of HCl = 0.0200 mol / 0.100 M = 0.200 L or 200 mL At this point, all HONH2 has reacted with HCl and has been converted to its conjugate acid HONH3+. 1. Calculate the new total volume after adding the HCl: Total volume = 100 mL (initial) + 200 mL (added) = 300 mL 2. Calculate the concentration of HONH3+ at equivalence point: \[ [\mathrm{HONH_3^+}] = \frac{0.0200}{0.300} \approx 0.0667 M \] 3. Write the dissociation of HONH3+: \[ \mathrm{HONH_3^+} \rightleftharpoons \mathrm{HONH_2} + \mathrm{H_3O^+} \] 4. Use the Kb value to find the Ka value: \[ K_a = \frac{K_w}{K_b} = \frac{1.0 \times 10^{-14}}{1.1 \times 10^{-8}} \approx 9.09 \times 10^{-7} \] 5. Set up an ICE table for HONH3+ dissociation: Let x = concentration of H3O+ formed \[ K_a = \frac{x^2}{0.0667 - x} \approx 9.09 \times 10^{-7} \] Solve for x ([H3O+]) and use it to calculate the pH, which will be approximately 4.14 at the equivalence point. #Solution#: e. Calculate the \(\mathrm{pH}\) after \(300.0 \mathrm{~mL}\) of \(\mathrm{HCl}\) has been added. Similar to the calculations in part b and c, determine the amount of HCl and weak base left after reaction, and calculate the pH after performing the necessary Kb calculations. The approximate pH after 300 mL of HCl has been added is 1.67. #Solution#: f. At what volume of \(\mathrm{HCl}\) added does the \(\mathrm{pH}=6.04\)? Similar to previous calculations, determine the volume of HCl needed. To do this we will need to use the pKa and equate the concentration between the weak base (HONH2) and its conjugate acid (HONH3+) using the Henderson-Hasselbalch equation. Find the moles of HONH2 and HONH3+ at pH = 6.04, calculate the moles of HCl needed, and use this to find the required volume of HCl. The volume of HCl needed to reach pH 6.04 is approximately 196.7 mL.