Respuesta :
Answer:
(a) Ka = 1.3 × 10⁻⁵
(b) Kb = 4.0 × 10⁻¹⁰
(c) Hydrazoic acid is the stronger acid.
(d) The propanate ion is a stronger base.
Explanation:
(a) the Ka value for propanoic acid (HC₂H₅CO₂)
Let's consider the acid reaction of propanoic acid in water.
HC₂H₅CO₂(aq) + H₂O(l) ⇄ C₂H₅CO₂⁻(aq) + H₃O⁺(aq)
HC₂H₅CO₂ is an acid and C₂H₅CO₂⁻ its conjugate base. The equilibrium constant for this reaction is Ka.
Now, let's consider the basic reaction of C₂H₅CO₂⁻.
C₂H₅CO₂⁻(aq) + H₂O(l) ⇄ HC₂H₅CO₂(aq) + OH⁻(aq)
The equilibrium constant for this reaction is Kb, where Kb = 7.8 × 10⁻¹⁰.
The relation between Ka and Kb is:
Ka × Kb = Kw = 1.0 × 10⁻¹⁴
Ka = Kw/Kb = 1.3 × 10⁻⁵
(b) the Kb value for the azide ion (N₃⁻)
Let's consider the basic reaction of the azide ion in water.
N₃⁻(aq) + H₂O(l) ⇄ HN₃(aq) + OH⁻(aq)
N₃⁻ is a base and HN₃ its conjugate acid. The equilibrium constant for this reaction is Kb.
Now, let's consider the acid reaction of HN₃.
HN₃(aq) + H₂O(l) ⇄ N₃⁻(aq) + H₃O⁺(aq)
The equilibrium constant for this reaction is Ka, where Ka = 2.5 × 10⁻⁵.
The relation between Ka and Kb is:
Ka × Kb = Kw = 1.0 × 10⁻¹⁴
Kb = Kw/Ka = 4.0 × 10⁻¹⁰
(c) the stronger acid between hydrazoic acid and propanoic acid
The higher the value of the Ka, the stronger the acid. Ka for hydrazoic acid is 2.5 × 10⁻⁵ and Ka for propanoic acid is 1.3 × 10⁻⁵, So, hydrazoic acid is the stronger acid.
(d) the stronger base between the azide ion and the propanoate ion
The higher the value of the Kb, the stronger the base. Kb for the azide ion is 4.0 × 10⁻¹⁰ and Kb for the propanate ion is 7.8 × 10⁻¹⁰, so the propanate ion is a stronger base.
The value of Ka for propanoic acid is 1.3 × 10⁻⁵ and value of Kb for azide ion is 4.0 × 10⁻¹⁰.
What is the relation between Ka & Kb?
Relation between acid dissociation constant (Ka) and base dissociation constant (Kb) is represents as:
Ka × Kb = 10⁻¹⁴
- Ka value for propanoic acid will be calculated by using the above formula and basic chemical equation for C₂H₅CO₂⁻ will be shown as:
C₂H₅CO₂⁻(aq) + H₂O(l) ⇄ HC₂H₅CO₂(aq) + OH⁻(aq)
Kb value for propanoate ion = 7.8 × 10⁻¹⁰(given)
Ka = 10⁻¹⁴ / 7.8 × 10⁻¹⁰ = 1.3 × 10⁻⁵
- Kb value for azide ion is also calculated by using the above formula and acid reaction of HN₃ will be shown as:
HN₃(aq) + H₂O(l) ⇄ N₃⁻(aq) + H₃O⁺(aq)
Ka value for azide ion (N₃⁻) = 2.5 × 10⁻⁵
Kb = 10⁻¹⁴ / 2.5 × 10⁻⁵ = 4.0 × 10⁻¹⁰
- Higher value of Ka & Kb will shows the stronger acid and base respectively. So hydrazoic acid is the stronger acid and propanate ion is a stronger base.
Hence value of Ka & Kb for propanoic acid and azide ion is 1.3×10⁻⁵ & 4.0×10⁻¹⁰ respectively, and hydrazoic is the stroger acid and propanate ion is a stronger base.
To know more about Ka & Kb, visit the below link:
https://brainly.com/question/26998
