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.
![### Calculating the Acid Dissociation Constant (Ka) of a Weak Acid
**Problem Statement:**
Calculate the Ka of a 0.35 M weak acid HA if the pH of the solution is 2.97.
*Assume the temperature is 25°C.*
**Solution:**
To find the acid dissociation constant (Ka), we need to use the information provided:
- The concentration of the weak acid (HA) = 0.35 M
- The pH of the solution = 2.97
**Step-by-Step Calculation:**
1. **Find the concentration of H⁺ ions:**
The pH of a solution is defined by the equation:
\[
\text{pH} = -\log [\text{H}^+]
\]
Rearranging this equation to solve for \([\text{H}^+]\):
\[
[\text{H}^+] = 10^{-\text{pH}}
\]
Substituting the given pH value:
\[
[\text{H}^+] = 10^{-2.97} \approx 1.07 \times 10^{-3} \text{ M}
\]
2. **Set up the expression for the acid dissociation constant (Ka):**
The dissociation of a weak acid (HA) in water is represented as:
\[
\text{HA} \rightleftharpoons \text{H}^+ + \text{A}^-
\]
The expression for \(K_a\) is:
\[
K_a = \frac{[\text{H}^+][\text{A}^-]}{[\text{HA}]}
\]
3. **Determine the equilibrium concentrations:**
- At equilibrium, the concentration of \([\text{H}^+]\) = \([\text{A}^-]\) = \(1.07 \times 10^{-3}\) M (since one mole of \(\text{HA}\) dissociates to give one mole of \(\text{H}^+\) and one mole of \(\text{A}^-)\).
- The initial concentration of \(\text{HA}\) was 0.35 M, and a small amount, \(1.07 \times](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ff65fb7b2-ba85-4e67-8fe1-01f97a564d1e%2F22af52b0-67d9-4d50-9822-188ed11c8d69%2Flgw1y2_processed.png&w=3840&q=75)
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