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
Ionic Equilibrium
Chemical equilibrium and ionic equilibrium are two major concepts in chemistry. Ionic equilibrium deals with the equilibrium involved in an ionization process while chemical equilibrium deals with the equilibrium during a chemical change. Ionic equilibrium is established between the ions and unionized species in a system. Understanding the concept of ionic equilibrium is very important to answer the questions related to certain chemical reactions in chemistry.
Arrhenius Acid
Arrhenius acid act as a good electrolyte as it dissociates to its respective ions in the aqueous solutions. Keeping it similar to the general acid properties, Arrhenius acid also neutralizes bases and turns litmus paper into red.
Bronsted Lowry Base In Inorganic Chemistry
Bronsted-Lowry base in inorganic chemistry is any chemical substance that can accept a proton from the other chemical substance it is reacting with.
**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
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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](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F277ae00b-d056-42ed-a729-386f4fca5973%2F02935c2b-281d-41b9-9b57-4db0a43ccd16%2Fto10p9b_processed.jpeg&w=3840&q=75)
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