Problem 40 - General Chemistry: Principles and Modern Applications

Question

Calculate the pH at the points in the titration of \(25.00 \mathrm{mL}\) of \(0.160 \mathrm{M} \mathrm{HCl}\) when (a) \(10.00 \mathrm{mL}\) and \((\mathrm{b}) 15.00 \mathrm{mL}\) of 0.242 M KOH have been added.

Step-by-Step Solution

Answer

The pH at the points in the titration of \(0.275 \, \mathrm{M} \, \mathrm{KOH}\) when (a) \(15.00 \, \mathrm{mL}\) and (b) \(20.00 \, \mathrm{mL}\) of \(0.350 \, \mathrm{M} \, \mathrm{HCl}\) have been added are calculated based on the formulas outlined above.

Step 1: Calculate the initial moles of KOH

This value is given by the formula \(\text{Molarity (M)} \times \text{Volume (L)} = \text{Moles}\).In this case, \(0.275 \, \mathrm{M} \times 0.020 \, \mathrm{L} = 0.0055 \, \mathrm{moles}\) of KOH.

Step 2: Calculate the moles of HCl added (Part a)

Again, we use the formula \(\text{Molarity (M)} \times \text{Volume (L)} = \text{Moles}\).For part (a), \(0.350 \, \mathrm{M} \times 0.015 \, \mathrm{L} = 0.00525 \, \mathrm{moles}\) of HCl.

Step 3: Determine nature of solution and calculate resulting pH (Part a)

Subtract moles of HCl from moles of KOH to see if the solution is acidic or basic.In this case, \(0.0055 - 0.00525 = 0.00025 \, \mathrm{moles}\) of KOH are left; the solution is basic.To find the pH, we first find the pOH by dividing the remaining moles of KOH by the total volume of the solution (in this case, \(20.00 + 15.00 = 35.00 \, \mathrm{ml}\) or \(0.035 \, \mathrm{L}\), then use \(\text{pOH} = -\log [\text{OH}^-]\) to find the pOH, and finally use \(\text{pH} + \text{pOH} = 14\) to find the pH.

Step 4: Calculate the moles of HCl added (Part b)

For part (b), we calculate moles of HCl as follows: \(0.350 \, \mathrm{M} \times 0.020 \, \mathrm{L} = 0.0070 \, \mathrm{moles}\) of HCl.

Step 5: Determine nature of solution and calculate resulting pH (Part b)

In this case, after we subtract moles of KOH from moles of HCl (\(0.00700 - 0.0055 = 0.0015 \, \mathrm{moles}\)), we see that the solution is acidic as we have excess \(\mathrm{HCl}\). Calculate the moles of H^+ ions left by dividing moles of HCl left by the total volume of the solution, which is now \(20.00 + 20.00 = 40.00 \, \mathrm{ml}\) or \(0.040 \, \mathrm{L}\). Use \(\text{pH} = -\log [\text{H}^+]\) to find the pH.

Key Concepts

pH calculation

Understanding how to calculate pH is crucial in acid-base titrations. The pH scale measures how acidic or basic a solution is, ranging from 0 to 14. To calculate pH, you need to know the concentration of hydrogen ions (H⁺) in the solution. When the solution is acidic, pH can be calculated using the formula:

• \( \text{pH} = -\log[\text{H}^+] \)

For basic solutions, it is helpful to first calculate pOH (the measure of hydroxide ions, OH^-) and then convert it to pH using the relationship:

• \( \text{pH} + \text{pOH} = 14 \)

In the titration exercise, accurately determining pH involves calculating the excess concentration of either H⁺ or OH^- after reaction completion and applying these formulas.

Moles of reactants

In titration, calculating the moles of reactants is a key step to understanding the chemical reaction occurring. When acid (HCl) and base (KOH) react, they neutralize each other. The number of moles can be calculated using the formula:

• Molarity (M) \( \times \) Volume (L) = Moles

For example, to find out how many moles of HCl or KOH you have in your solution, multiply the concentration (molarity) of your solution by the volume used.
During a titration, you want to find out how much of the acid or base has reacted. In the given problem, you will calculate it initially for one reactant and then use it to identify the condition of the solution post-reaction.

Acidic and basic solutions

Solutions can either be acidic or basic depending on the concentration of H⁺ and OH^- ions.
- **Acidic solutions** have a higher concentration of H⁺ ions compared to OH^- ions. Consequently, their pH value is less than 7. - **Basic solutions** have a higher concentration of OH^- ions and a pH value greater than 7. In the context of a titration, the nature of the solution depends on the quantity of remaining unreacted acid or base. For instance, if more of the base (KOH) remains unreacted, the solution becomes basic, and vice versa for an acidic solution.

Neutralization reaction

A neutralization reaction occurs when an acid and a base react to form water and a salt, effectively neutralizing each other. This is a fundamental concept in acid-base titrations. When you add an acid to a base (or the other way around), they react to produce a neutral product if mixed in the correct proportions. By measuring how much of one reactant is required to completely react with another, you can calculate the equivalence point - where the amount of acid equals the amount of base. The completion of a neutralization reaction can be seen analytically by the change in pH.
In titration problems, identifying the point at which all acid has reacted with the base (or vice versa) helps to determine the extent of neutralization and remaining pH of the solution.