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)\( by 0.200\)M \mathrm{HNO}_{3}\( . Calculate the \)\mathrm{pH}$ of the resulting solution after the following volumes of \(\mathrm{HNO}_{3}\) have been added. $$ \begin{array}{ll}{\text { a. } 0.0 \mathrm{mL}} & {\text { d. } 40.0 \mathrm{mL}} \\ {\text { b. } 20.0 \mathrm{mL}} & {\text { e. } 50.0 \mathrm{mL}} \\ {\text { c. } 25.0 \mathrm{mL}} & {\text { f. } 100.0 \mathrm{mL}}\end{array} $$

The balanced reaction equation is: \[H_2NNH_2 + HNO_3 → H_2NNH_3^+ + NO_3^-\]

Initially, we have the following: - H2NNH2 = (100.0 mL)(0.100 M) = 10.0 mmol - HNO3 = not present - H2NNH3+ = not present - NO3- = not present Now, let's find the moles of the species present and pH after each volume of HNO3:

No HNO3 is added yet, so we only have the initial base, H2NNH2, present. The base will hydrolyze with water as follows: \[H_2NNH_2 + H_2O \rightleftharpoons OH^- + H_2NNH_3^+\] Now, we can use the Kb to find the concentration of OH-: \[K_b = \frac{[OH^-][H_2NNH_3^+]}{[H_2NNH_2]} = 3.0 \times 10^{-6}\] Let x be the concentration of OH-. Now we have: \[(3.0 \times 10^{-6}) = \frac{x^2}{(0.100-x)}\] Solving for x, we get [OH-] = x ≈ 5.48 × 10-4 M. Finally, we can determine the pH: \[pH = 14 - pOH = 14 - (-\log{[OH^-]}) \approx 14 - 3.26 = 10.74\]

Moles of HNO3 = (20.0 mL)(0.200 M) = 4.0 mmol Since 4.0 mmol of HNO3 reacts with the H2NNH2, we have: - H2NNH2 = 10.0 mmol - 4.0 mmol = 6.0 mmol - HNO3 = 0 - H2NNH3+ = 4.0 mmol - NO3- = 4.0 mmol Now, we use the Henderson-Hasselbalch equation: \[pH = pK_b + \log{\frac{[base]}{[acid]}}\] Plug in the values, remembering that conjugate base = [H2NNH2], and acid = [H2NNH3+]: \[pH \approx 5.52 + \log{\frac{6}{4}} \approx 5.52 + 0.18 = 5.70\]

Moles of HNO3 = (25.0 mL)(0.200 M) = 5.0 mmol Since 5.0 mmol of HNO3 reacts with the H2NNH2, we have: - H2NNH2 = 10.0 mmol - 5.0 mmol = 5.0 mmol - HNO3 = 0 - H2NNH3+ = 5.0 mmol - NO3- = 5.0 mmol Use the Henderson-Hasselbalch equation: \[pH = pK_b + \log{\frac{[base]}{[acid]}}\] Plug in the values: \[pH \approx 5.52 + \log{\frac{5}{5}} = 5.52 + 0 = 5.52\]

We will follow similar steps for the remaining volumes of HNO3: d. 40.0 mL: - Moles of HNO3 = 8.0 mmol - H2NNH2 = 2.0 mmol - H2NNH3+= 8.0 mmol \[pH \approx 5.52 + \log{\frac{2}{8}} \approx 5.52 - 0.60 = 4.92\] e. 50.0 mL: - Moles of HNO3 = 10.0 mmol - H2NNH2 = 0 mmol - H2NNH3+ = 10.0 mmol \[pH \approx 5.52 + \log{\frac{0}{10}}\] Since the ratio is undefined, we cannot use the Henderson-Hasselbalch equation. At this point, all of the base ('H2NNH2') has been neutralized by the HNO3. Because there will be some leftover HNO3, it would be better to calculate the acid concentration and use that to determine the pH. In this case, the pH would be approximately 3.76. f. 100.0 mL: - Moles of HNO3 = 20.0 mmol - All H2NNH2 is already neutralized (from part 'e') - pH would depend on the concentration of excess HNO3, which would be [H+] = [HNO3] = 0.200 M × (100.0 mL - 50.0 mL) ÷ (100.0 mL + 100.0 mL) = 0.050 M. The pH would be approximately 1.30. The resulting pH values after adding each volume of HNO3 are: a. 0.0 mL: 10.74 b. 20.0 mL: 5.70 c. 25.0 mL: 5.52 d. 40.0 mL: 4.92 e. 50.0 mL: 3.76 f. 100.0 mL: 1.30