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.
![### Chemical Equilibrium Constants
Understanding the equilibrium constants (K) for chemical reactions is crucial for predicting the direction of reactions and the concentrations of reactants and products at equilibrium. Below are two example reactions with given equilibrium constants.
1. **Reaction I:**
- **Equation:** \( \text{HF(aq)} \rightleftharpoons \text{H}^+(\text{aq}) + \text{F}^-(\text{aq}) \)
- **Equilibrium Constant (K):** \( 6.8 \times 10^{-4} \)
2. **Reaction II:**
- **Equation:** \( \text{H}_2\text{C}_2\text{O}_4(\text{aq}) \rightleftharpoons 2\text{H}^+(\text{aq}) + \text{C}_2\text{O}_4^{2-} \)
- **Equilibrium Constant (K):** \( 3.8 \times 10^{-6} \)
### Combined Reaction
The problem involves finding the equilibrium constant (K) for the combined reaction:
\[
2\text{HF(aq)} + \text{C}_2\text{O}_4^{2-}\text{(aq)} \rightleftharpoons 2\text{F}^-(\text{aq)} + \text{H}_2\text{C}_2\text{O}_4(\text{aq)}
\]
To determine the K value for this new reaction, we use the given equilibrium constants. For reactions that involve summing two or more elementary reactions, the equilibrium constant of the overall reaction is the product of the equilibrium constants of the individual steps.
#### Solution Method:
- Multiply the given K values for the elementary reactions.
### Calculation:
Let's denote:
- \(K_1 = 6.8 \times 10^{-4}\)
- \(K_2 = 3.8 \times 10^{-6}\)
Then:
\[
K = K_1 \times K_2 = (6.8 \times 10^{-4}) \times (3.8 \times 10^{-6})
\]
### Final K Value
To simplify the expression, perform the arithmetic operations:
- Multiply the coefficients: \( 6.8 \times 3.8 = 25.84](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ff93c0e8e-c503-4238-bd00-8ed92b9568ca%2F291c9d33-268c-4e44-b6e1-689ee3183005%2Fib5y6cd_processed.jpeg&w=3840&q=75)
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