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 \(0.200 M\) HNO \(_{3}\). Calculate the 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. \(100.0 \mathrm{mL}\)

Short answer

The pH of the solution at each point is as follows: a. 11.21 b. 10.53 c. 9.52 d. 1.84 e. 7.00 f. 1.30

Step-by-step solution

a. Initial solution (0.0mL HNO3)

Before titration, all we have in the solution is H2NNH2, a weak base. To calculate the pH, we will first find the pOH of the solution using Kb and an ICE table for the given reaction: \[ H_{2}NNH_{2}(aq) + H_{2}O(l) \rightleftharpoons H_{2}NNH_{3}^{+}(aq) + OH^{-}(aq) \] Initial concentrations: [H2NNH2] = 0.100 M [OH-] = 0 [H2NNH3+] = 0 Change in concentrations: [H2NNH2] = -x [OH-] = +x [H2NNH3+] = +x Equilibrium concentrations: [H2NNH2] = 0.100-x M [OH-] = x M [H2NNH3+] = x M Using Kb in the expression: \(K_b = \frac{{[H2NNH3+][OH-]}}{{[H2NNH2]}} = \frac{x^2}{0.100-x}\) Now solve for x: \(3.0*10^{-6} = \frac{x^2}{0.100-x}\) Solving for x, we get x = \(6.14*10^{-4} \) Thus, [OH-] = \(6.14*10^{-4} \) M Now find pOH: \( pOH = -log([OH^{-}]) = -log(6.14*10^{-4}) \) Therefore, pH = 14 - pOH

b. During titration (20.0mL HNO3)

When 20.0mL of 0.200M HNO3 is added, it reacts with the H2NNH2 in the solution. Calculate moles of HNO3 added and the reaction with H2NNH2: Moles of HNO3 = 0.020L * 0.200M = 0.0040mol Initial moles of H2NNH2 = 0.100M * 0.100L = 0.0100mol Moles left after reaction: H2NNH2 = 0.0100 - 0.0040 = 0.0060mol HNO3 = 0 H2NNH3+ = 0.0040mol New concentrations after reaction: [H2NNH2] = 0.0060mol / 0.120L = 0.0500M [H2NNH3+] = 0.0040mol / 0.120L = 0.0333M pOH = pKb + log(\(\frac{[H2NNH3+]}{[H2NNH2]}\)) pOH = -log(3.0*10^{-6}) + log(\(\frac{0.0333}{0.0500}\)) pH = 14 - pOH

c. At equivalence (25.0mL HNO3)

At equivalence, exactly all of the weak base reacts with the strong acid. Calculate moles of HNO3 added, moles of H2NNH2 leftover, and moles of H2NNH3+ produced: Moles of HNO3 = 0.025L * 0.200M = 0.0050mol Moles of H2NNH2 left = 0.0100mol - 0.0050mol = 0 Moles of H2NNH3+ = 0.0050mol [H2NNH3+] = 0.0050mol/0.125L = 0.0400M Since the equivalence point is reached, we have a salt (H2NNH3+). Calculate the pH of the salt, formed from a weak base and a strong acid: pOH = pKw - pKa pOH = 14 - (-log(3.0 * 10^{-6})) pH = 14 - pOH

d. During titration (40.0mL HNO3)

After adding 40.0mL of 0.200M HNO3, calculate moles of HNO3 added, moles of H2NNH2 left, moles of H2NNH3+ produced and moles of HNO3 left: Moles of HNO3 added = 0.040L * 0.200M = 0.0080mol [H2NNH2] = 0 Moles of HNO3 left = 0.0080mol - 0.0100mol = -0.0020mol [HNO3] = -0.0020mol/0.140L = 0.0143M At this point in the titration, enough HNO3 has been added to react with all the H2NNH2 and there is some HNO3 left. Since HNO3 is a strong acid, calculate the pH as follows: pH = -log([H3O+]) pH = -log(0.0143)

e. After titration (50.0mL HNO3)

After adding 50.0mL of 0.200M HNO3, calculate moles of HNO3 added, moles of H2NNH2 left, moles of H2NNH3+ produced and moles of HNO3 left: Moles of HNO3 added = 0.050L * 0.200M = 0.0100mol [H2NNH2] = 0 Moles of HNO3 left = 0.0100mol - 0.0100mol = 0 [HNO3] = 0 Since all the weak base has reacted and no HNO3 is left, we have a neutral solution: pH = 7

f. After titration (100.0mL HNO3)

After adding 100.0mL of 0.200M HNO3, calculate moles of HNO3 added, moles of H2NNH2 left, moles of H2NNH3+ produced and moles of HNO3 left: Moles of HNO3 added = 0.100L * 0.200M = 0.0200mol [H2NNH2] = 0 Moles of HNO3 left = 0.0200mol - 0.0100mol = 0.0100mol [HNO3] = 0.0100mol/0.200L = 0.0500M Calculate the pH as follows: pH = -log([H3O+]) pH = -log(0.0500)

Key concepts

pH Calculation

When dealing with acid-base titrations, calculating pH is crucial. For a weak base like hydrazine, \( \text{H}_2\text{NNH}_2 \), reacting with a strong acid such as nitric acid \( \text{HNO}_3 \), the pH changes as titrant is added.

Initially, you need to find the pOH derived from the weak base. Here's how:

• Set up an ICE (Initial, Change, Equilibrium) table for the dissociation of the base in water.
• Use the base dissociation constant \( K_b \) to find the equilibrium concentrations.
• Calculate \([\text{OH}^-]\), then use the formula \( \text{pOH} = -\log([\text{OH}^-]) \).

Since pH + pOH = 14, simply subtract pOH from 14 to find the pH.

Understanding these calculations ensures that you can accurately determine the pH at various titration points.

Equilibrium Concentration

Equilibrium concentration is essential for understanding chemical reactions during titration. In the initial state before any acid is added, only the weak base \( \text{H}_2\text{NNH}_2 \) exists.

To find the equilibrium concentrations:

• Set initial concentrations for all species. \([\text{H}_2\text{NNH}_2] = 0.100\, \text{M}\), others = 0.
• Predict changes that occur, as base dissociates into \( \text{H}_2\text{NNH}_3^+ \) and \( \text{OH}^- \).
• Use \( K_b \) to establish the relationship \( \frac{x^2}{0.100-x} = 3.0 \times 10^{-6} \).

Solving this equation reveals \([\text{OH}^-]\) at equilibrium, key for pH determination.

This careful tracking of concentrations helps in understanding reaction dynamics at each stage of titration.

Weak Base and Strong Acid Reaction

In titrations with a weak base and a strong acid, like \( \text{H}_2\text{NNH}_2 \) and \( \text{HNO}_3 \), understanding the reactions between them is vital.

As acid is added:

• The strong acid \( \text{HNO}_3 \) completely dissociates and reacts with the weak base.
• During this reaction, the base is converted to \( \text{H}_2\text{NNH}_3^+ \), forming its conjugate acid.
• The point where moles of acid equals moles of base is the equivalence point.

At the equivalence point, no base remains, leaving only the salt \( \text{H}_2\text{NNH}_3^+ \), which can influence the pH due to its acidic nature.

By analyzing these reaction stages, you can predict how the pH and concentrations change throughout the titration process.