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.
![**Neutralization Calculation: Determining the Volume of HCl Needed to Neutralize NaOH**
**Problem Statement:**
How many milliliters (ml) of 0.25 M HCl is required to neutralize 50.0 ml of 0.30 M NaOH?
**Detailed Explanation:**
To solve this problem, you need to use the concept of neutralization in chemistry. Neutralization occurs when an acid and a base react to form water and a salt. The number of moles of H\(^+\) ions from the acid will equal the number of moles of OH\(^-\) ions from the base.
The relationship between the concentrations and volumes of the acid and base can be given by the formula:
\[ \text{M}_1 \cdot \text{V}_1 = \text{M}_2 \cdot \text{V}_2 \]
where:
- \( \text{M}_1 \) is the molarity of the acid (HCl)
- \( \text{V}_1 \) is the volume of the acid
- \( \text{M}_2 \) is the molarity of the base (NaOH)
- \( \text{V}_2 \) is the volume of the base
**Given:**
- \( \text{M}_1 = 0.25 \) M
- \( \text{M}_2 = 0.30 \) M
- \( \text{V}_2 = 50.0 \) ml
**Calculation:**
\[ 0.25 \, \text{M} \times \text{V}_1 = 0.30 \, \text{M} \times 50.0 \, \text{ml} \]
\[ \text{V}_1 = \frac{0.30 \, \text{M} \times 50.0 \, \text{ml}}{0.25 \, \text{M}} \]
\[ \text{V}_1 = \frac{15}{0.25} \]
\[ \text{V}_1 = 60 \, \text{ml} \]
**Conclusion:**
You would need 60.0 ml of 0.25 M HCl to neutralize 50.0 ml of 0.30 M NaOH.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F2d4c5a9e-ae57-4a9b-bcac-2cfb7874c09d%2F6595be40-65e7-46f9-bdec-2a9c240a38b9%2Fce5cw5_processed.jpeg&w=3840&q=75)
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