Given that K₂ for HF is 6.3 × 10-4 at 25 °C, what is the value of K₁ for Fat 25 °C? Kb = Given that K for (CH3)²N is 6.3 × 10−5 at 25 °C, what is the value of Kå for (CH3)²NH+ at 25 °C? Ka =
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.
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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.
![### Chemistry Problem: Calculating Equilibrium Constants
In this exercise, we explore the relationship between acid dissociation constants (\(K_a\)) and base dissociation constants (\(K_b\)) at 25 °C.
#### Problem 1:
Given that the \(K_a\) for HF is \(6.3 \times 10^{-4}\) at 25 °C, what is the value of \(K_b\) for F\(^-\) at 25 °C?
\[ K_b = \text{[Text Box for Answer]} \]
#### Problem 2:
Given that \(K_b\) for (CH\(_3\))\(_3\)N is \(6.3 \times 10^{-5}\) at 25 °C, what is the value of \(K_a\) for (CH\(_3\))\(_3\)NH\(^+\) at 25 °C?
\[ K_a = \text{[Text Box for Answer]} \]
### Concept Explanation
The relationship between \(K_a\) and \(K_b\) for a conjugate acid-base pair is given by the equation:
\[ K_a \times K_b = K_w \]
Where \(K_w\) is the ion-product constant of water at 25 °C, equal to \(1.0 \times 10^{-14}\).
### Solution Steps
1. **For HF and F\(^-\):**
- Use the equation \(K_a \times K_b = K_w\).
- Rearrange to find \(K_b\):
\[ K_b = \frac{K_w}{K_a} \]
2. **For (CH\(_3\))\(_3\)N and (CH\(_3\))\(_3\)NH\(^+\):**
- Again, use the formula \(K_a \times K_b = K_w\).
- Rearrange to find \(K_a\):
\[ K_a = \frac{K_w}{K_b} \]
Plug in the given values to find the unknown constants.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F7a6c3af6-91f1-478b-99db-a0b0fd9518f2%2F6a7b1ce2-98d8-48b8-bd35-e306d32762c9%2Fxyvs4f_processed.png&w=3840&q=75)
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