Question

Consider the titration of \(25.0 \mathrm{~mL}\) of \(0.125 \mathrm{M} \mathrm{HCl}\) with \(0.100 \mathrm{M}\) KOH. Calculate the pH after the addition of each of the following volumes of base: (a) \(3.0 \mathrm{~mL}\) (b) \(20 \mathrm{~mL}\) (c) \(65 \mathrm{~mL}\)

Short answer

(a) pH ≈ 1.00, (b) pH ≈ 1.60, (c) pH ≈ 12.57.

Step-by-step solution

Calculate Initial Moles of HCl

We start by calculating the initial moles of HCl present in the solution. The concentration of HCl is given by \( 0.125 \mathrm{M} \) and the volume is \( 25.0 \mathrm{~mL} \): \[ \text{Moles of } \mathrm{HCl} = 0.125 \cancel{\mathrm{~M}} \times \frac{25.0 \cancel{\mathrm{~mL}}}{1000} = 0.003125 \text{ moles} \]

Case (a) - Calculate pH after Adding 3.0 mL of KOH

1. **Calculate Moles of KOH Added**: 0.100 M KOH and 3.0 mL added. \[ \text{Moles of } \mathrm{KOH} = 0.100 \cancel{\mathrm{~M}} \times \frac{3.0 \cancel{\mathrm{~mL}}}{1000} = 0.0003 \text{ moles} \] 2. **Determine Remaining Moles of HCl**:\[ 0.003125 - 0.0003 = 0.002825 \text{ moles of } \mathrm{HCl} \] 3. **Calculate New [H⁺] Concentration**: Total volume = 25.0 mL + 3.0 mL = 28.0 mL: \[ [\mathrm{H}^+] = \frac{0.002825 \cancel{\text{ moles}}}{0.0280 \cancel{\text{ L}}} = 0.10089 \mathrm{M} \] 4. **Convert to pH**: \[ \text{pH} = -\log(0.10089) \approx 1.00 \]

Case (b) - Calculate pH after Adding 20 mL of KOH

1. **Calculate Moles of KOH Added**: 0.100 M KOH and 20 mL added. \[ \text{Moles of } \mathrm{KOH} = 0.100 \cancel{\mathrm{~M}} \times \frac{20.0 \cancel{\mathrm{~mL}}}{1000} = 0.002 \text{ moles} \] 2. **Determine Remaining Moles of HCl**:\[ 0.003125 - 0.002 = 0.001125 \text{ moles of } \mathrm{HCl} \] 3. **Calculate New [H⁺] Concentration**: Total volume = 25.0 mL + 20 mL = 45.0 mL: \[ [\mathrm{H}^+] = \frac{0.001125 \cancel{\text{ moles}}}{0.0450 \cancel{\text{ L}}} = 0.025 \mathrm{M} \] 4. **Convert to pH**: \[ \text{pH} = -\log(0.025) \approx 1.60 \]

Case (c) - Calculate pH after Adding 65 mL of KOH

1. **Calculate Moles of KOH Added**: 0.100 M KOH and 65 mL added. \[ \text{Moles of } \mathrm{KOH} = 0.100 \cancel{\mathrm{~M}} \times \frac{65.0 \cancel{\mathrm{~mL}}}{1000} = 0.0065 \text{ moles} \]2. **Determine Excess Moles of KOH**: All HCl neutralized and some excess KOH remains: \[ 0.0065 - 0.003125 = 0.003375 \text{ moles of KOH} \] 3. **Calculate New [OH⁻] Concentration**: Total volume = 25.0 mL + 65 mL = 90.0 mL: \[ [\mathrm{OH}^-] = \frac{0.003375 \cancel{\text{ moles}}}{0.0900 \cancel{\text{ L}}} = 0.0375 \mathrm{M} \] 4. **Convert [OH⁻] to pH**: - Calculate pOH: \[ \text{pOH} = -\log(0.0375) \approx 1.43 \] - Use \( \text{pH} + \text{pOH} = 14 \): \[ \text{pH} = 14 - 1.43 = 12.57 \]

Conclusion: Summary of pH Values

- (a) After adding 3.0 mL of KOH, the pH is approximately 1.00. - (b) After adding 20 mL of KOH, the pH is approximately 1.60. - (c) After adding 65 mL of KOH, the pH is approximately 12.57.

Key concepts

pH calculation

pH calculation is an important concept in chemistry that helps us determine the acidity or basicity of a solution. The pH value is derived from the concentration of hydrogen ions (\(\text{H}^+\)) in the solution. For acidic solutions, like \(\text{HCl}\), the pH is found using the formula:

• \[ \text{pH} = -\log[\text{H}^+] \]

This formula converts the concentration of hydrogen ions into a more usable measure of acidity. In the context of a titration exercise, the pH value changes as more base is added to an acid. Initially, the solution is acidic, and the pH is lower. As neutralization occurs, the pH rises. Eventually, if excess base is added, the solution can become basic. Calculating pH is essential in determining the end-point of a titration and understanding how the balance between an acid and a base shifts during the process.
Understanding how pH changes during a titration will help you comprehend how acids and bases interact and foresee the impact of adding different volumes of titrant.

neutralization reaction

A neutralization reaction is a chemical process where an acid reacts with a base to produce water and a salt. This reaction is fundamental in acid-base chemistry and is central to titration experiments. In our exercise, \(\text{HCl}\), an acid, is neutralized by \(\text{KOH}\), a base. When these two react, the following general reaction occurs:

• \[ \text{HCl (aq)} + \text{KOH (aq)} \rightarrow \text{KCl (aq)} + \text{H}_2\text{O (l)} \]

This equation shows how hydrogen ions from the acid combine with hydroxide ions (\(\text{OH}^-\)) from the base to form water. The leftover ions from each substance combine to form a neutral salt, in this case, potassium chloride (\(\text{KCl}\)).
During a titration, the point at which the acid and base fully react with each other is known as the equivalence point. Past this point, the solution changes from primarily acidic to primarily basic if base is in excess. Understanding neutralization is key to mastering titration techniques and predicting the behavior of solutions during experiments.

moles calculation

Calculating moles is a crucial step in many chemical reactions, including titrations. Moles provide a way to quantify the number of molecules or atoms participating in a reaction. In the titration exercise, we calculated moles for both \(\text{HCl}\) and \(\text{KOH}\) to determine how much of each reactant is present. The formula for calculating moles from concentration and volume is:

• \[ \text{Moles} = \text{Concentration (M)} \times \text{Volume (L)} \]

In this exercise, we converted milliliters to liters by dividing by 1000. Using this calculation, we understood how many moles of \(\text{HCl}\) were initially in the solution and how many moles of \(\text{KOH}\) were added during the titration.
Calculating moles helps in balancing chemical reactions and determining when complete neutralization is achieved. This step is imperative in understanding not only how much of a reactant remains but also if any excess exists after a reaction is complete.

stoichiometry

Stoichiometry involves using the balanced chemical equation to determine the relationships between reactants and products in a reaction. This is pivotal in performing titration calculations accurately. In the acid-base reaction of \(\text{HCl}\) with \(\text{KOH}\), stoichiometry helps us understand that one mole of \(\text{HCl}\) reacts exactly with one mole of \(\text{KOH}\) to form one mole of \(\text{KCl}\) and one mole of water. This one-to-one ratio simplifies calculations and allows us to predict the outcome of mixing different amounts of reactants.
By understanding stoichiometry, you can determine how much of each substance is required to completely react with another and predict the amounts of product formed. It is a powerful tool that gives chemists the ability to manipulate chemical reactions and make accurate predictions about the results of reactions.
In our exercise scenario, stoichiometry guides the calculation process, ensuring we accurately assess the remaining reactants and resulting pH levels throughout the different stages of the titration.