Problem 39
Question
Write an equation that describes the equilibrium that exists when nitric acid dissolves in water. Identify each of the four species in solution as either Bronsted acids or Bronsted bases. Does the equilibrium favor the products or the reactants?
Step-by-Step Solution
Verified Answer
The equilibrium favors the products, with HNO₃ as a Brønsted acid and NO₃⁻ as a base.
1Step 1: Write the Dissolution Reaction
When nitric acid (HNO₃) dissolves in water, it donates a proton (H⁺) to water, producing hydronium ions (H₃O⁺) and nitrate ions (NO₃⁻). The equilibrium equation for this reaction is: \[\text{HNO₃} + \text{H₂O} \rightleftharpoons \text{H₃O}^+ + \text{NO₃}^-\] This reaction shows nitric acid dissociating into hydronium ions and nitrate ions.
2Step 2: Identify the Species as Brønsted Acids or Bases
In the equation, HNO₃ acts as a Brønsted acid because it donates a proton (H⁺) to water. Water (
H₂O) accepts the proton to become H₃O⁺ and acts as a Brønsted base. H₃O⁺, the hydronium ion, is the conjugate acid that forms, while NO₃⁻ is the conjugate base that results from the deprotonation of HNO₃.
3Step 3: Determine the Equilibrium Direction
Nitric acid is a strong acid, meaning it ionizes completely in water. This implies that the equilibrium lies far to the right, favoring the formation of products (H₃O⁺ and NO₃⁻) over the reactants (HNO₃ and H₂O). Therefore, the equilibrium significantly favors the product side.
Key Concepts
Brønsted AcidBrønsted BaseEquilibrium Direction
Brønsted Acid
When discussing the concept of a Brønsted acid, we are focusing on a substance that donates a proton, which is essentially a hydrogen ion (H⁺), in a chemical reaction.
In the dissolution of nitric acid (HNO₃) in water, HNO₃ acts as the Brønsted acid, since it donates a proton to water.
In the dissolution of nitric acid (HNO₃) in water, HNO₃ acts as the Brønsted acid, since it donates a proton to water.
- The formula for this reaction is: \[\text{HNO₃} + \text{H₂O} \rightleftharpoons \text{H₃O}^+ + \text{NO₃}^-\]
- After donating a proton, HNO₃ becomes NO₃⁻, the nitrate ion.
- This transformation demonstrates the principal behavior of a Brønsted acid—proton donation.
Brønsted Base
While the Brønsted acid in a reaction is the donor of
H⁺
, a Brønsted base, in contrast, accepts a hydrogen ion. In the example of nitric acid dissolving in water, water (
H₂O
) functions as the Brønsted base.
This is because it accepts the hydrogen ion from HNO₃ and forms H₃O⁺ , the hydronium ion.
This is because it accepts the hydrogen ion from HNO₃ and forms H₃O⁺ , the hydronium ion.
- During this process, H₂O acts as a proton acceptor.
- This proton acceptance transforms water into H₃O⁺ , linking the conversion to its role as a base.
- The ability of a substance to accept a proton defines it as a Brønsted base regardless of its original state.
Equilibrium Direction
In chemical reactions, equilibrium direction refers to the tendency of a reaction to favor either the reactants or products once the reaction has reached a stable state. This concept is vital in understanding how strong acids like nitric acid behave in solution.
Since HNO₃ is a strong acid, it dissociates almost completely in water.
This has a significant impact on the equilibrium of the reaction.
Since HNO₃ is a strong acid, it dissociates almost completely in water.
This has a significant impact on the equilibrium of the reaction.
- In this scenario: \[\text{HNO₃} + \text{H₂O} \rightleftharpoons \text{H₃O}^+ + \text{NO₃}^-\]
- The equilibrium heavily favors the right side, meaning a majority of HNO₃ converts to H₃O⁺ and NO₃⁻.
- This is indicative of the strong acid’s propensity to release its protons almost completely.
Other exercises in this chapter
Problem 37
Write a balanced equation for the reaction of barium hydroxide with nitric acid.
View solution Problem 38
Write a balanced equation for the reaction of aluminum hydroxide with sulfuric acid.
View solution Problem 40
Write an equation that describes the equilibrium that exists when the weak acid benzoic acid \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2} \mathrm{H}\ri
View solution Problem 41
Write two chemical equations, one that shows \(\mathrm{H}_{2} \mathrm{O}\) reacting (with HBr) as a Bronsted base and a second that shows \(\mathrm{H}_{2} \math
View solution