Question

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

Short answer

a. The initial pH is 8.68. b. The pH after 25.0 mL of HCl has been added is 8.39. c. The pH after 70.0 mL of HCl has been added is 4.74. d. The pH at the equivalence point is 5.36. e. The pH after 300.0 mL of HCl has been added is 1.52. f. The volume of HCl needed to reach a pH of 6.04 is approximately 58.9 mL.

Step-by-step solution

Initial pH calculation

To find the initial pH, we only need to consider the dissociation of HONH2, since no HCl has been added yet. The equilibrium expression is: \(HONH_2 + H_2O \leftrightarrow HONH_3^+ + OH^-\) Using the Kb expression: \(K_b = \frac{[HONH_3^+][OH^-]}{[HONH_2]}\) Since the initial concentration of HONH2 is given as 0.200 M, we will set up an ICE table and solve for x (the change in concentration of the species): \(K_b = \frac{x^2}{0.200 - x}\) Plug in the given Kb value: \(1.1\times10^{-8} = \frac{x^2}{0.200 - x}\) Solve for x (concentration of OH-) and use this to calculate the pH using the formula: \(pOH = -log_{10}([OH^-])\) and \(pH = 14 - pOH\) b. Calculate the pH after 25.0 mL of HCl has been added.

pH after HCl addition

First, we need to find out the new concentration of the base (HONH2) by calculating the moles and concentration of added HCl (0.100 M x 0.025 L). Then, set up a new ICE table with updated initial concentrations and calculate the pH as in part a. c. Calculate the pH after 70.0 mL of HCl has been added.

pH after HCl addition

Repeat the same procedure as in part b, using the new volume of added HCl (0.070 L). d. Calculate the pH at the equivalence point.

pH at equivalence point

At the equivalence point, all HONH2 has been neutralized by HCl. Calculate the volume of HCl needed to neutralize HONH2 completely (100.0 mL x 0.200 M = x L x 0.100 M). Use this volume to calculate the concentration of the HONH3+ ion formed at the equivalence point, and find the pH using the Ka expression for the HONH3+ ion (Ka = Kw / Kb for the conjugate acid-base pair). e. Calculate the pH after 300.0 mL of HCl has been added.

pH after HCl addition

Repeat the procedure as in parts b and c, using the new volume of added HCl (0.300 L). f. At what volume of HCl added does the pH = 6.04?

Volume of HCl for specific pH

To find the volume of HCl needed to reach a pH of 6.04, set up an ICE table for a reaction between HONH2 and the HCl added up to that point. The initial concentrations will depend on the moles of each species. Use moles of HCl (moles_HCl = moles_HONH_erased) to find the volume of HCl needed at this point by dividing moles_HCl by the concentration of HCl (0.100 M).

Key concepts

pH calculation

Calculating pH is an essential step in understanding acid-base chemistry during titration processes. The pH of a solution indicates its acidity or alkalinity. To find the pH, we generally follow these steps:

• Identify the concentrations of hydronium ions
• Use the formula: \( \text{pH} = -\log_{10}[\text{H}^+] \)

The pH scale ranges from 0 to 14, where 0 represents a strong acid, 14 represents a strong base, and 7 is neutral.In the context of titration, as an acid or base is added, the pH changes correspondingly. This is crucial for calculating at different volumes of titrant added to reach the complete neutralization called the equivalence point.Understanding the influence of added titrant allows for accurate predictions of pH levels over the various stages of the titration process.

acid-base reaction

Acid-base reactions are a fundamental concept in chemistry and are characterized by the exchange of protons between reactants. In a typical titration problem, you are dealing with an acid reacting with a base. For this exercise, the reaction involves HONH\(_2\) (a weak base) and HCl (a strong acid).

• Acids: Donate protons (H\(^+\))
• Bases: Accept protons

The reaction in this exercise can be represented as:\[\text{HONH}_2 + \text{HCl} \rightarrow \text{HONH}_3^+ + \text{Cl}^- \]Each molecule of acid donates a proton to the base, forming conjugate acid and base pairs.It is important to recognize that, in a strong acid-weak base reaction like this, the base will be entirely converted to its conjugate acid at the equivalence point.Understanding this exchange of protons helps predict changes in pH throughout the titration, ultimately leading us to interesting outcomes at different stages of the addition of HCl.

ICE table

An ICE table is a helpful tool in equilibrium calculations, allowing you to organize information before, during, and after the reaction to predict concentrations of reactants and products.ICE stands for Initial, Change, and Equilibrium, and it sets up a structured manner to calculate unknown values. Here's how it works in our titration:

• Initial: Start with known initial concentrations
• Change: Determine the changes as the reaction proceeds
• Equilibrium: Calculate concentrations at equilibrium

For a base like HONH\(_2\), you set up the initial concentration and identify the changes due to the reaction with HCl. The ICE table helps determine how much each species is "consumed" or "produced" as the equilibrium is established. In practice, when \( \text{HONH}_2 \) reacts with HCl, the changes calculated in the ICE table can directly lead to solving the equilibrium expression for pH.

equivalence point

The equivalence point in a titration is when the number of moles of titrant added equals the number of moles of the substance being titrated. This is a crucial stage because it signifies the completion of the neutralization reaction. In the context of titration between a weak base (HONH\(_2\)) and a strong acid (HCl), the equivalence point can be particularly insightful.At the equivalence point:

• All original moles of the base have been completely neutralized by the acid
• The solution will primarily consist of the conjugate acid and accompanying ions

The pH can still change dramatically here. Since a strong acid is involved, you would often find the pH to be slightly acidic.The pH at this point is determined by the hydrolysis of the conjugate acid.Understanding the equivalence point is key to mastering titration and predicting the pH at this critical juncture in the reaction.