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**
**Question:**
What volume (in mL) of 0.2600 M HBr is required to neutralize 50.00 mL of 0.7000 M KOH?
**Explanation:**
This question involves a neutralization reaction where hydrobromic acid (HBr) reacts with potassium hydroxide (KOH). To solve for the volume of HBr needed, you can use the equation for neutralization:
\( M_1 \cdot V_1 = M_2 \cdot V_2 \)
Where:
- \( M_1 \) and \( V_1 \) are the molarity and volume of the acid (HBr).
- \( M_2 \) and \( V_2 \) are the molarity and volume of the base (KOH).
1. **Known Values:**
- \( M_2 = 0.7000 \, \text{M} \) (KOH)
- \( V_2 = 50.00 \, \text{mL} \) (KOH)
- \( M_1 = 0.2600 \, \text{M} \) (HBr)
2. **Calculate \( V_1 \):**
Using the formula:
\[
V_1 = \frac{(M_2 \cdot V_2)}{M_1}
\]
Plug in the values:
\[
V_1 = \frac{(0.7000 \, \text{M} \cdot 50.00 \, \text{mL})}{0.2600 \, \text{M}}
\]
Calculate:
\[
V_1 = \frac{35.00}{0.2600} = 134.615 \, \text{mL}
\]
3. **Conclusion:**
Therefore, 134.615 mL of 0.2600 M HBr is required to neutralize 50.00 mL of 0.7000 M KOH.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F81a44892-e725-436e-9c8e-21f599a0ba84%2F6a85de04-1774-4b94-85c8-01caf9f0bf9b%2Fq4e86j.jpeg&w=3840&q=75)
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