Question

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 \(\mathrm{pH}=6.04\) ?

Short answer

a. The initial pH of the HONH2 solution is approximately 10.17. b. The pH after the addition of 25.0 mL of HCl is approximately 9.90. c. The pH after the addition of 70.0 mL of HCl is approximately 9.10. d. The pH at the equivalence point is approximately 4.14. e. The pH after the addition of 300.0 mL of HCl is approximately 1.67. f. The volume of HCl needed to reach a pH of 6.04 is approximately 196.7 mL.

Step-by-step solution

Initial pH of the HONH2 solution

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.

pH after addition of 25.0 mL HCl

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.

pH at 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.

Key concepts

Equivalence Point

In an acid-base titration, the equivalence point is a crucial stage where the amount of titrant added corresponds exactly to the amount needed to completely react with the solution being titrated. This means that, at the equivalence point, the moles of acid equal the moles of base in the solution.

For the titration of the weak base hydroxylamine (\( \text{HONH}_2 \)) with hydrochloric acid (HCl), the equivalence point can be calculated once the moles of each solution are equal.

• Initially, 0.0200 moles of \( \text{HONH}_2 \) are present.
• So, 0.0200 moles of HCl are required to reach the equivalence point, which translates to 200 mL of 0.100 M HCl.

At the equivalence point, \( \text{HONH}_2 \) is fully converted into its conjugate acid, \( \text{HONH}_3^+ \). The pH at this stage will not be neutral (pH 7) because \( \text{HONH}_3^+ \) has acidic characteristics, contributing to an acidic solution.

Henderson-Hasselbalch Equation

The Henderson-Hasselbalch Equation is a very useful formula to find the pH of a buffer solution, which contains a weak acid and its conjugate base or a weak base and its conjugate acid. It simplifies the calculation by relating the pH to the \(\text{pK}_a\) (or \(\text{pK}_b\)) and the concentration ratio of the species.

For our example of \( \text{HONH}_2 \) and its conjugate acid \( \text{HONH}_3^+ \), the equation is written as:\[\text{pH} = \text{pK}_a + \log\left(\frac{[\text{HONH}_2]}{[\text{HONH}_3^+]}\right)\]

Here, \( \text{pK}_a \) can be derived from \( \text{K}_b \) using the formula:\[\text{pK}_a = 14 - \text{pK}_b\]

This equation helps find the pH at any point before or after reaching the equivalence point in a titration if there is a buffer present. It's particularly crucial when determining the pH at 196.7 mL of HCl addition, where pH = 6.04.

Weak Base

A weak base is a substance that does not completely dissociate into its ions in solution. In this case, hydroxylamine (\( \text{HONH}_2 \)) is a weak base, meaning it only partially ionizes in water, producing \( \text{HONH}_3^+ \) and \( \text{OH}^- \) ions. This partial ionization is why weak base solutions typically have a higher pH than strong acids yet are not extremely basic.

The dissociation is represented by the equilibrium:\[\text{HONH}_2 \rightleftharpoons \text{HONH}_3^+ + \text{OH}^-\]

The degree of dissociation is indicated by \( \text{K}_b \), the base dissociation constant, which is quite small. A low \( \text{K}_b \) value, such as \( 1.1 \times 10^{-8} \), emphasizes the minimal dissociation and lower strength of \( \text{HONH}_2 \) as a base.

In a titration, as you add a strong acid like HCl, the base continues to react, forming \( \text{HONH}_3^+ \), which plays a key role in establishing the final pH values at each stage of titration.

pH Calculations

In titrations involving weak bases and strong acids, pH calculations are essential to understanding the solution's acidity at any given point. Calculating the pH involves several steps:

• Initial pH: For the initial solution of the weak base \( \text{HONH}_2 \), calculate the pH using the \( \text{K}_b \) value to find \( [\text{OH}^-] \), then convert that into \( \text{pOH} \) and finally into pH.
• Intermediate pH: After adding specific volumes of HCl, compute the concentration of \( \text{HONH}_2 \) left and use it to find the pH, employing the equilibrium equation derived from \( \text{K}_b \).
• Equivalence Point pH: Use the conversion of the base to its conjugate acid to compute pH, involving calculation of \( \text{K}_a \) from \( \text{K}_b \), and then finding \( [\text{H}_3\text{O}^+] \) to determine the pH.

Accurate calculations reveal how the solution transitions from a basic environment with \( \text{HONH}_2 \) to a more acidic environment as the titration progresses, especially past the equivalence point.