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.
![**Understanding Percent Ionization of Ammonia**
**Question:**
What is the percent ionization of a 0.0075 M solution of ammonia?
**Given:**
\( K_b(\text{NH}_3) = 1.8 \times 10^{-5} \)
**Solution:**
To calculate the percent ionization, follow these steps:
1. **Write the equilibrium expression for the ionization of ammonia:**
\[
\text{NH}_3 + \text{H}_2\text{O} \rightleftharpoons \text{NH}_4^+ + \text{OH}^-
\]
2. **Set up the expression for the base ionization constant (\(K_b\)):**
\[
K_b = \frac{[\text{NH}_4^+][\text{OH}^-]}{[\text{NH}_3]}
\]
3. **Assume a small degree of ionization (x):**
Let \(x\) be the change in concentration.
\[
0.0075 - x \approx 0.0075
\]
4. **Plug in the values into the \(K_b\) expression:**
\[
1.8 \times 10^{-5} = \frac{x^2}{0.0075}
\]
5. **Solve for \(x\):**
\[
x^2 = 1.8 \times 10^{-5} \times 0.0075
\]
\[
x^2 = 1.35 \times 10^{-7}
\]
\[
x = \sqrt{1.35 \times 10^{-7}}
\]
\[
x \approx 3.67 \times 10^{-4}
\]
6. **Calculate the percent ionization:**
\[
\text{Percent ionization} = \left( \frac{x}{0.0075} \right) \times 100\%
\]
\[
\text{Percent ionization} = \left( \frac{3.67 \times 10^{-4}}{0.0075} \right) \times 100
\]
\[
\text{Percent ionization} \approx](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa4c65346-a6bd-47be-a891-a239e86f6fd1%2F7ecf9579-e477-41c3-a0a7-d9d1cb8f2341%2F2kf1hdr_processed.jpeg&w=3840&q=75)
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