Question

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. $$ \begin{array}{ll}{\text { a. } 0.0 \mathrm{mL}} & {\text { d. } 40.0 \mathrm{mL}} \\ {\text { b. } 20.0 \mathrm{mL}} & {\text { e. } 50.0 \mathrm{mL}} \\ {\text { c. } 25.0 \mathrm{mL}} & {\text { f. } 100.0 \mathrm{mL}}\end{array} $$

Short answer

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

Step-by-step solution

Balance the equation

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

Calculate the moles of the species before titration

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:

a. 0.0 mL of HNO3 added

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: \[pH = 14 - pOH = 14 - (-\log{[OH^-]}) \approx 14 - 3.26 = 10.74\]

b. 20.0 mL of HNO3 added

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+]: \[pH \approx 5.52 + \log{\frac{6}{4}} \approx 5.52 + 0.18 = 5.70\]

c. 25.0 mL of HNO3 added

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: \[pH \approx 5.52 + \log{\frac{5}{5}} = 5.52 + 0 = 5.52\]

Steps d, e, and f

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

Key concepts

pH calculation

The pH is an important measure of the acidity or basicity of a solution. It is calculated using the concentration of hydrogen ions \([H^+]\), where pH is defined as \[pH = -\log{[H^+]}\].
In the context of titrations, calculating pH can be subtle as it involves various species. Here are key points:

• Strong acids and bases fully dissociate in water, leading to straightforward pH calculations.
• Weak acids and bases, like \(\mathrm{H}_2\mathrm{NNH}_2\), partially dissociate, requiring expressions involving equilibrium constants such as \(K_a\) or \(K_b\).
• During a titration, as an acid or base is added, the solution's pH changes depending on the remaining reactants and products.

Understanding these principles allows the calculation of pH at any point during a titration, considering neutralization reactions and concentrations of resulting ions.

Henderson-Hasselbalch equation

The Henderson-Hasselbalch equation is a robust tool in calculating the pH of solutions during a titration. It is particularly useful for buffer solutions, which contain a weak acid and its conjugate base, or a weak base and its conjugate acid. The equation is expressed as:\[pH = pK_a + \log{\frac{[\text{base}]}{[\text{acid}]}}\]In our example problem involving \(\mathrm{H}_2\mathrm{NNH}_2\) and \(\mathrm{HNO}_3\), the equation helps find pH when a weak base reacts with a strong acid, forming a weak acid. Here’s what happens:

• As \(\mathrm{HNO}_3\) is added, \(\mathrm{H}_2\mathrm{NNH}_2\) reacts to form \(\mathrm{H}_2\mathrm{NNH}_3^+\), a conjugate acid.
• The equation then calculates pH using \(pK_b\) equivalently, since the reaction converts base to acid.
• It's particularly practical when the base and the conjugate acid are close in concentration, simplifying the logarithmic calculations.

Remember, the equation is most accurate when you are within the buffer's optimal pH range, around the \(pK_a\) or \(pK_b\) value.

neutralization reaction

Neutralization reactions are fundamental in acid-base chemistry. During a titration, neutralization occurs when acids and bases react to form water and salt.
Here's how it works:

• A characteristic of these reactions is that they proceed until the moles of \[H^+\]\ from the acid equal those of \[OH^-\]\ from the base, achieving neutrality.
• The balanced equation in our exercise shows \(H_2NNH_2 + HNO_3 \to H_2NNH_3^+ + NO_3^-\).
• At complete neutralization, contained \(H_2NNH_2\) is entirely converted to its conjugate acid by \(\mathrm{HNO}_3\), and no excess base remains.
• The titration has endpoints distinguished by a significant pH change, indicated by calculated pH transitions as seen at 50 mL and 100 mL.

Each increment of titrant affects the concentration of the resulting ions, carefully altering the solution's pH. Understanding this helps in accurately tracking the progress and endpoints of titrations.