Answer:
pH = 5.47
Explanation:
The equilibrium that takes place is:
HIO ↔ H⁺ + IO⁻
Ka =
= 2.3 * 10⁻¹¹
At equilibrium:
<u>Replacing those values in the equation for Ka and solving for x:</u>

Then [H⁺]=3.39 * 10⁻⁶, thus pH = 5.47
Answer:
Kb = [CH₃NH₃⁺] × [OH⁻] / [CH₃NH₂]
Explanation:
According to Brönsted-Lowry acid-base theory:
- An acid is a substance that donates H⁺.
- A base is a substance that accepts H⁺.
When methylamine reacts with water, it behaves as a Brönsted-Lowry base, according to the following reaction.
CH₃NH₂(aq) + H₂O(l) ⇄ CH₃NH₃⁺(aq) + OH⁻(aq)
The basic equilibrium constant (Kb) is:
Kb = [CH₃NH₃⁺] × [OH⁻] / [CH₃NH₂]
Answer:
c. Compound 2 is more acidic because its conjugate base is more resonance stabilized
Explanation:
You haven't told us what the compounds are, so let's assume that the formula of Compound 1 is HCOCH₂OH and that of Compound 2 is CH₃COOH.
The conjugate base of 2 is CH₃COO⁻. It has two important resonance contributors, and the negative charge is evenly distributed between the two oxygen atoms.
CH₃COOH + H₂O ⇌ CH₃COO⁻ + H₃O⁺
The stabilization of the conjugate base pulls the position of equilibrium to the right, so the compound is more acidic than 1.
The smallest halogen atom is fluorine
C
This is because 10+5=15
15/45=0.3