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.
![**Question:**
What is the pH of a 3.40 M solution of HNO₂ (Kₐ = 4.5 x 10⁻⁴)?
**Response Section:**
- Your Answer: [Text Box]
This question requires you to calculate the pH of a nitrous acid solution using the given concentration and acid dissociation constant (Kₐ).
**Instructions for Solving:**
1. Write the dissociation equation for nitrous acid (HNO₂):
\[
\text{HNO}_2 \rightleftharpoons \text{H}^+ + \text{NO}_2^-
\]
2. Set up the expression for the acid dissociation constant (Kₐ):
\[
Kₐ = \frac{[\text{H}^+][\text{NO}_2^-]}{[\text{HNO}_2]}
\]
where:
- [H⁺] = concentration of hydrogen ions
- [NO₂⁻] = concentration of nitrite ions
- [HNO₂] = initial concentration minus the change in concentration x due to dissociation
3. Use the given concentration of HNO₂ to determine changes in concentration:
- Initial: [HNO₂] = 3.40 M, [H⁺] = 0, [NO₂⁻] = 0
- Change: [HNO₂] = 3.40 - x, [H⁺] = x, [NO₂⁻] = x
4. Assume x is much less than the initial concentration (3.40 M), simplifying the Kₐ equation:
\[
4.5 \times 10^{-4} = \frac{x^2}{3.40}
\]
5. Solve for x to find [H⁺]:
- x = [H⁺] = √(3.40 * 4.5 x 10⁻⁴)
6. Calculate the pH:
\[
\text{pH} = -\log[\text{H}^+]
\]
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