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 pH of a 0.25 M NH₃ Solution
**Problem Statement:**
What is the pH of a 0.25 M NH₃ solution?
Given that \( K_b = 1.8 \times 10^{-5} \).
**Solution:**
To calculate the pH of an ammonia (NH₃) solution, follow these steps:
1. **Calculate \( [OH⁻] \):**
Use the equilibrium expression for ammonia dissociation in water:
\[ NH₃(aq) + H₂O(l) \rightleftharpoons NH₄⁺(aq) + OH⁻(aq) \]
The \( K_b \) expression is:
\[ K_b = \frac{[NH₄⁺][OH⁻]}{[NH₃]} \]
For this solution:
\[ 1.8 \times 10^{-5} = \frac{x^2}{0.25 - x} \]
Where \( x \) represents the concentration of \( OH⁻ \) ions.
2. **Approximate \( x \) for Small \( K_b \):**
Since \( K_b \) is small, assume \( x \) is much less than 0.25:
\[ 1.8 \times 10^{-5} \approx \frac{x^2}{0.25} \]
\[ x^2 = 1.8 \times 10^{-5} \times 0.25 \]
\[ x^2 = 4.5 \times 10^{-6} \]
\[ x = \sqrt{4.5 \times 10^{-6}} \]
\[ x \approx 2.12 \times 10^{-3} \]
Thus, \( [OH⁻] \approx 2.12 \times 10^{-3} \) M.
3. **Find \( pOH \):**
\[ pOH = -\log[OH⁻] \]
\[ pOH = -\log(2.12 \times 10^{-3}) \]
\[ pOH \approx 2.67 \]
4. **Calculate pH:**
Using the relation between pH and pOH:
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