Question

Consider the titration of \(100.0 \mathrm{mL}\) of \(0.200 \mathrm{M}\) HONH \(_{2}\) by \(0.100 \mathrm{M}\) HCl. \(\left(K_{\mathrm{b}} \text { for } \mathrm{HONH}_{2}=1.1 \times 10^{-8} .\right)\) a. Calculate the \(\mathrm{pH}\) after \(0.0 \mathrm{mL}\) of HCI has been added. b. Calculate the \(\mathrm{pH}\) after \(25.0 \mathrm{mL}\) of HCl has been added. c. Calculate the \(\mathrm{pH}\) after \(70.0 \mathrm{mL}\) of HCl has been added. d. Calculate the \(\mathrm{pH}\) at the equivalence point. e. Calculate the \(\mathrm{pH}\) after \(300.0 \mathrm{mL}\) of HCl has been added. f. At what volume of HCl added does the \(\mathrm{pH}=6.04 ?\)

Short answer

a. pH = 11.12 b. pH = 10.81 c. pH = 10.11 d. pH = 9.27 e. pH = 1.52 f. Volume of HCl added = 53.0 mL

Step-by-step solution

Write the reaction

The reaction between HONH\(_2\) and HCl is: HONH\(_2\) + HCl → HONH\(_3^+\) + Cl\(^-\) Step 2: Calculating moles of HONH\(_2\)

Calculate moles of HONH\(_2\)

We have 100.0 mL of 0.200 M HONH\(_2\). Therefore, the moles of HONH\(_2\) are: Moles of HONH\(_2\) = (Volume × Concentration) = (100.0 mL × 0.200 M) = 0.0200 mol HONH\(_2\) Step 3: Calculating moles of HCl at each stage

Calculate moles of HCl

Use the volume of HCl that was added at each stage of the titration to calculate the moles of HCl: Moles of HCl = (Volume × Concentration) a. 0.0 mL of 0.100 M HCl = 0.000 mol HCl b. 25.0 mL of 0.100 M HCl = 0.00250 mol HCl c. 70.0 mL of 0.100 M HCl = 0.00700 mol HCl d. At the equivalence point, equal moles of HONH\(_2\) and HCl will have reacted, so 0.0200 mol HCl is added. e. 300.0 mL of 0.100 M HCl = 0.0300 mol HCl Step 4: Determine species present for each stage and calculate pH

Determine species and calculate pH

For each stage, calculate the remaining moles of HONH\(_2\), HCl, and the formed HONH\(_3^+\), and use the Kb value (1.1 × 10\(^{-8}\)) to find the pH. a. 0.0 mL HCl added, buffer solution: HONH\(_2\) and HONH\(_3^+\) (with 0 added HCl) will be present, use Kb value to calculate pH. Consider HONH\(_2\) as the base and HCl as the acid. pH = 14 - pOH pOH = -log[\(OH^-\)] Kb = [HONH\(_3^+\)][OH\(^-\)] / [HONH\(_2\)] 1.1 × 10\(^{-8}\) = [\(x\)][\(x\)] / [0.200 - \(x\)] b. 25.0 mL HCl added, buffer solution: 0.00250 mol HCl reacts with HONH\(_2\), and we have excess HONH\(_2\). Use Kb value, the moles of HONH\(_2\), and HONH\(_3^+\) to calculate pH. c. 70.0 mL HCl added, buffer solution: 0.00700 mol HCl reacts with HONH\(_2\), and we have excess HONH\(_2\). Use Kb value, the moles of HONH\(_2\), and HONH\(_3^+\) to calculate pH. d. Equivalence point, 0.0200 mol HCl added: We have only HONH\(_3^+\) (a weak acid) and Cl\(^-\). Use the Kb value and water dissociation constant, Kw (1.0 × 10\(^{-14}\)), to calculate Ka for HONH\(_3^+\): Ka = Kw / Kb Calculate the concentration of HONH\(_3^+\) in the solution and use the Ka value to find the pH. e. 300.0 mL HCl added, acidic solution: Excess HCl is present, use the remaining moles of HCl and total volume to find the H\(^+\) concentration and calculate the pH directly. f. pH = 6.04, find volume of HCl added: Set the pH and use Kb value and moles of HONH\(_2\) and HONH\(_3^+\) to find the moles of HCl added. Then, use the concentration of HCl to find the volume of HCl added.

Key concepts

Acid-Base Reaction

Understanding acid-base reactions is vital, especially in titrations. In the given exercise, we're dealing with a weak base (HONH\(_2\)) reacting with a strong acid (HCl). The reaction between these substances forms an acid (HONH\(_3^+\)) and Cl\(^{-}\). This process is simply an acid-base neutralization reaction, where the base accepts protons (H\(^+\)) from the acid.

• The chemical equation looks like this: HONH\(_2\) + HCl → HONH\(_3^+\) + Cl\(^-\)
• Such reactions are common in titration setups, where a reactant is added gradually to measure the unknown concentration of a solution.
• This forms a solution that can be acidic or alkaline depending on the relative strengths of the acid and base involved.

For weak bases, like HONH\(_2\), the resulting solution will shift in pH towards acidity upon addition of strong acids. The understanding of these reactions helps us predict the shifts in pH during titration.

Equivalence Point

An equivalence point in titration is a pivotal moment where the amount of acid equals the amount of base in the solution. In this exercise, it's the point where the moles of HONH\(_2\) equal the moles of added HCl.

• This occurs typically without a distinct pH, varying based on the strengths of the acids and bases involved.
• For a weak base reacting with a strong acid, the pH at the equivalence point will tend to be in the acidic range due to the formation of a weak acid (HONH\(_3^+\)).
• It is calculated using the formula: Total moles of base = Total moles of acid.

This stage is crucial for determining the point in titration where complete neutralization has occurred, but it is not necessarily the end of a pH calculation of a titration. Subsequent additions of the titrant can further influence pH.

pH Calculation

Calculating pH in titration is a critical skill. The pH tells us how acidic or basic a solution is at any point during the titration process. In this specific case, we'll deal with different scenarios of how much HCl is added to HONH\(_2\).

• Initially, the solution contains 0.200 M of HONH\(_2\) (a base), and no HCl. The pH is determined using the base dissociation constant \(K_b\).
• As HCl is gradually added, the pH drops because HONH\(_2\) and HCl react to form HONH\(_3^+\) and Cl\(^-\).
• To calculate pH, you have to understand where the solution stands in the titration process: before, at, or after the equivalence point.

By calculating the concentrations of different species (like OH\(^-\) or H\(^+\)), you can determine the pH using formulas like: pOH = -log[OH\(^-\)] and pH = 14 - pOH. At each step of adding HCl, the overall balance between HONH\(_2\), HCl, and HONH\(_3^+\) changes, affecting the pH.

Weak Base Titration

Weak base titrations involve a weak base, like HONH\(_2\), reacting with a strong acid. In this exercise, we observe the changes in pH at different stages of adding HCl. Weak bases have unique properties that affect their titration behavior.

• Unlike strong bases, weak bases partially ionize in water. This partial ionization affects how the pH curves during titration.
• During titration, the weak base reacts with the strong acid to form a conjugate acid, which creates a buffer region before reaching the equivalence point.
• At and after the equivalence point, the solution may become acidic, since the strong acid dictates the basicity of the mixture.

Understanding these properties is essential for performing accurate pH calculations and for predicting the behavior of the weak base throughout the titration. By knowing the \(K_b\) of HONH\(_2\), we can better estimate the pH changes as more HCl is added.