Consider the titration of \(100.0 \mathrm{~mL}\) of \(0.100 \mathrm{M} \mathrm{H}_{2} \mathrm{NNH}_{2}\) \(\left(K_{\mathrm{b}}=3.0 \times 10^{-6}\right)\) by \(0.200 \mathrm{M} \mathrm{HNO}_{3} .\) Calculate the \(\mathrm{pH}\) of the resulting solution after the following volumes of \(\mathrm{HNO}_{3}\) have been added. a. \(0.0 \mathrm{~mL}\) b. \(20.0 \mathrm{~mL}\) c. \(25.0 \mathrm{~mL}\) d. \(40.0 \mathrm{~mL}\) e. \(50.0 \mathrm{~mL}\) f. \(100.0 \mathrm{~mL}\)

- Volumes: \(100.0 \ \text{mL} \text{ H}_{2}\text{NNH}_{2},\) volumes of \(\text{HNO}_{3}\) are given in the problem. - Concentrations: \(0.100 \ \text{M H}_{2}\text{NNH}_{2}, 0.200 \ \text{M HNO}_{3}\) - Equilibrium constant: \(K_\text{b}=3.0 \times 10^{-6}\)

- Moles of base, \(\text{H}_{2}\text{NNH}_{2}: moles_{base} = initial\ base\ concentration \times initial\ base\ volume - \frac{acid\ volume \times acid\ concentration}{2}\) - Moles of acid, \(\text{HNO}_3: moles_{acid} = acid\ volume \times acid\ concentration\)

For each volume, calculate the moles of species and then the resulting concentrations. The moles and concentrations of the species will be different at each volume. a. \(0.0\ \text{mL HNO}_3\) b. \(20.0\ \text{mL HNO}_3\) c. \(25.0\ \text{mL HNO}_3\) d. \(40.0\ \text{mL HNO}_3\) e. \(50.0\ \text{mL HNO}_3\) f. \(100.0\ \text{mL HNO}_3\)

For each case, calculate the pH using the appropriate formulas and relations between the equilibrium constant and the concentrations of the species present. a. The added acid volume is zero, so the solution is just the base with no acid. Calculate the initial pOH using the Kb value and then find the pH by subtracting it from 14. b. and c. This is the buffer region, use the Henderson-Hasselbalch equation to calculate pH: \[pH = pK_a + \log{\frac{[\text{A}^-]}{[\text{HA}]}}\] d. This is the equivalence point, the base and acid have reacted completely. Calculate the pH using only the conjugate species that are produced in the reaction. e. and f. This is the excess acid region, calculate pH directly by taking the negative logarithm of the H+/H3O+ ion concentration in the solution.