Ammonia (NH) is an example of a Brønsted-Lowry Base.

. Ammonia (NH
) is an example of a Brønsted-Lowry Base. i. Define the Brønsted-Lowry acid-base theory. ii. What’s the pH of a solution of ammonia that has a concentration of 0.335 M? The K
of ammonia is
. First complete the ICE chart. HA H
O
A
Initial Change Equilibrium iii. Calculate the pH. Please show all work.

The Correct Answer and Explanation is:

Here is the solution to the chemistry problem.

11. Ammonia (NH₃) is an example of a Brønsted-Lowry Base.

i. Define the Brønsted-Lowry acid-base theory.
The Brønsted-Lowry theory defines acids and bases based on the transfer of protons (H⁺ ions).

• A Brønsted-Lowry acid is a substance that donates a proton (H⁺).
• A Brønsted-Lowry base is a substance that accepts a proton (H⁺).

ii. What’s the pH of a solution of ammonia that has a concentration of 0.335 M? The K₋ of ammonia is 1.8 × 10⁻⁵. First complete the ICE chart.

The problem involves ammonia (NH₃), a weak base, reacting with water. The chemical equation is:
NH₃(aq) + H₂O(l) ⇌ NH₄⁺(aq) + OH⁻(aq)

Note: The ICE chart provided in the question has column headers (HA, H₃O⁺, A⁻) that are appropriate for a generic acid dissociation. For this problem about the base ammonia, the correct species are the reactant (NH₃), the conjugate acid (NH₄⁺), and the hydroxide ion (OH⁻). The chart below is completed with the correct values for the ammonia reaction, corresponding to the species NH₃, NH₄⁺, and OH⁻.

Here is the completed ICE chart for the reaction of ammonia in water:

iii. Calculate the pH. Please show all work.

Step 1: Write the base dissociation constant (K₋) expression.
The expression for K₋ is based on the equilibrium concentrations from the ICE chart.
K₋ = [NH₄⁺][OH⁻] / [NH₃]

Step 2: Substitute the equilibrium values into the K₋ expression.
1.8 × 10⁻⁵ = (x)(x) / (0.335 – x)

Step 3: Simplify and solve for x.
Since K₋ is very small, we can assume that x will be much smaller than the initial concentration of 0.335 M. Therefore, we can approximate (0.335 – x) ≈ 0.335.
(To check this assumption: [NH₃]initial / K₋ = 0.335 / (1.8 × 10⁻⁵) ≈ 18611, which is much greater than 100, so the approximation is valid).

1.8 × 10⁻⁵ = x² / 0.335
x² = (1.8 × 10⁻⁵) × 0.335
x² = 6.03 × 10⁻⁶
x = √(6.03 × 10⁻⁶)
x = 2.456 × 10⁻³

Step 4: Calculate the pOH.
From the ICE chart, x represents the equilibrium concentration of hydroxide ions, [OH⁻].
[OH⁻] = 2.456 × 10⁻³ M

pOH = -log[OH⁻]
pOH = -log(2.456 × 10⁻³)
pOH ≈ 2.61

Step 5: Calculate the pH.
The relationship between pH and pOH at 25°C is pH + pOH = 14.
pH = 14 – pOH
pH = 14 – 2.61
pH = 11.39