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 Acidity Constant (Ka) of Hypobromous Acid
#### Problem Statement:
The pH of a 0.064-M solution of hypobromous acid (HOBR but usually written HBrO) is 4.97. Calculate \( K_a \).
#### Solution:
Given data:
- The concentration of hypobromous acid (HBrO): 0.064 M
- The pH of the solution: 4.97
To find:
- The acidity constant \( K_a \)
First, we need to find the concentration of hydrogen ions \([H^+]\) using the pH value.
The pH is related to the hydrogen ion concentration by the formula:
\[ \text{pH} = -\log[H^+] \]
Rearranging the formula to find \([H^+]\):
\[ [H^+] = 10^{-\text{pH}} \]
Substituting the given pH value (4.97):
\[ [H^+] = 10^{-4.97} \]
Now calculate \([H^+]\):
\[ [H^+] \approx 1.07 \times 10^{-5} \, \text{M} \]
Next, we setup the expression for the dissociation of hypobromous acid:
\[ \text{HBrO} \leftrightharpoons \text{H}^+ + \text{BrO}^- \]
At equilibrium:
- The initial concentration of HBrO is 0.064 M.
- The change in concentration is \( x \) for both \([H^+]\) and \([\text{BrO}^-]\).
Thus,
\[ [H^+] = [\text{BrO}^-] = 1.07 \times 10^{-5} \]
\[ [\text{HBrO}] = 0.064 - 1.07 \times 10^{-5} \]
Since \( 1.07 \times 10^{-5} \) is very small compared to 0.064, we approximate:
\[ [\text{HBrO}] \approx 0.064 \, \text{M} \]
The expression for \( K_a \) of HBrO is:
\[ K_a = \frac{[H^+][\text{BrO](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F5fe141b1-9817-4d6d-9b09-948fe13c457c%2F1f2b83ed-1abc-4307-aaad-45c0eaddc651%2Ffomwlul_processed.jpeg&w=3840&q=75)
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