The diagram below represents a monoprotic acid, HA. If 14.75 mL 0.0200 M KOH are added to 32.47 mL 0.0100 M solution of the above acid, which is the final pH? O4.08 4.23 2.08 3.08

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### Question 7

The diagram below represents a monoprotic acid, HA.

![Diagram of monoprotic acid](link-to-image)

**Description of Diagram:** 
The diagram shows a beaker with a liquid containing two types of particles: larger red spheres and smaller blue spheres. The larger red spheres likely represent the undissociated form of the monoprotic acid (HA), while the smaller blue spheres represent either the hydrogen ions (H⁺) or the conjugate base (A⁻) formed upon dissociation.

**Problem Statement:**
If 14.75 mL of 0.0200 M KOH is added to 32.47 mL of 0.0100 M solution of the above acid, what is the final pH?

**Answer Choices:**
- 4.08
- 4.23
- 2.08
- 3.08

---

Here's a step-by-step explanation for finding the pH after the addition of KOH to the monoprotic acid solution:

1. Calculate the moles of KOH added:
   \[
   \text{Moles of KOH} = 14.75 \text{ mL} \times \left( \frac{0.0200 \text{ moles}}{1000 \text{ mL}} \right) = 0.000295 \text{ moles}
   \]

2. Calculate the moles of the acid (HA) initially present:
   \[
   \text{Moles of HA} = 32.47 \text{ mL} \times \left( \frac{0.0100 \text{ moles}}{1000 \text{ mL}} \right) = 0.0003247 \text{ moles}
   \]

3. Since KOH is a strong base, it will completely neutralize the acid. Determine the remaining moles of HA and the moles of A⁻ produced:
   \[
   \text{Moles of HA after reaction} = 0.0003247 \text{ moles} - 0.000295 \text{ moles} = 0.0000297 \text{ moles}
   \]
   \[
   \text{Moles of A⁻ formed} = 0.000295 \text{ moles}
   \]

4. Determine
Transcribed Image Text:### Question 7 The diagram below represents a monoprotic acid, HA. ![Diagram of monoprotic acid](link-to-image) **Description of Diagram:** The diagram shows a beaker with a liquid containing two types of particles: larger red spheres and smaller blue spheres. The larger red spheres likely represent the undissociated form of the monoprotic acid (HA), while the smaller blue spheres represent either the hydrogen ions (H⁺) or the conjugate base (A⁻) formed upon dissociation. **Problem Statement:** If 14.75 mL of 0.0200 M KOH is added to 32.47 mL of 0.0100 M solution of the above acid, what is the final pH? **Answer Choices:** - 4.08 - 4.23 - 2.08 - 3.08 --- Here's a step-by-step explanation for finding the pH after the addition of KOH to the monoprotic acid solution: 1. Calculate the moles of KOH added: \[ \text{Moles of KOH} = 14.75 \text{ mL} \times \left( \frac{0.0200 \text{ moles}}{1000 \text{ mL}} \right) = 0.000295 \text{ moles} \] 2. Calculate the moles of the acid (HA) initially present: \[ \text{Moles of HA} = 32.47 \text{ mL} \times \left( \frac{0.0100 \text{ moles}}{1000 \text{ mL}} \right) = 0.0003247 \text{ moles} \] 3. Since KOH is a strong base, it will completely neutralize the acid. Determine the remaining moles of HA and the moles of A⁻ produced: \[ \text{Moles of HA after reaction} = 0.0003247 \text{ moles} - 0.000295 \text{ moles} = 0.0000297 \text{ moles} \] \[ \text{Moles of A⁻ formed} = 0.000295 \text{ moles} \] 4. Determine
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