Achemist prepared aqueous solutions of two weak acids: solution a is hydrazoic acid (hn3), and solution b is propanoic acid (hc2h5co2). at 25°c, the acid-dissociation constant (ka) for hydrazoic acid is 2.5 ✕ 10−5 and the base-dissociation constant (kb) for the propanoate ion (c2h5co2−) is 7.8 ✕ 10−10. determine the following. (a) the ka value for propanoic acid (hc2h5co2) (b) the kb value for the azide ion (n3−) (c) the stronger acid between hydrazoic acid and propanoic acid hydrazoic acid propanoic acid (d) the stronger base between the azide ion and the propanoate ion azide ion

(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.

i'm assuming that it's c.) red bank..

your answer is a. the enthalpy of formation for a pure element in its standard state is always positive.