What is the pH of a 0.250 M solution of NH4Br, given that the Ka of NH4+ is 5.6 x 10^-10?
The Correct Answer and Explanation is :
To calculate the pH of a 0.250 M solution of ammonium bromide (NH₄Br), we first need to understand the dissociation and behavior of the ammonium ion (NH₄⁺) in water. Since ammonium bromide is a salt formed from the weak base ammonia (NH₃) and a strong acid (HBr), it dissociates in water to produce NH₄⁺ and Br⁻ ions. The Br⁻ ion, being the conjugate base of a strong acid, does not affect the pH, while NH₄⁺ is acidic and will affect the pH of the solution.
Step 1: Write the dissociation equation for NH₄⁺
Ammonium ion (NH₄⁺) is a weak acid and partially dissociates in water:
[
NH₄⁺ \rightleftharpoons NH₃ + H⁺
]
Step 2: Set up the equilibrium expression
The dissociation of NH₄⁺ can be represented by the acid dissociation constant (Ka) expression:
[
K_a = \frac{[NH₃][H⁺]}{[NH₄⁺]}
]
We are given that the Ka for NH₄⁺ is (5.6 \times 10^{-10}), and the initial concentration of NH₄⁺ is 0.250 M.
Step 3: Set up an ICE table
Let (x) represent the concentration of H⁺ ions that dissociate from NH₄⁺:
| Species | Initial (M) | Change (M) | Equilibrium (M) |
|---|---|---|---|
| NH₄⁺ | 0.250 | -x | 0.250 – x |
| NH₃ | 0 | +x | x |
| H⁺ | 0 | +x | x |
Step 4: Substitute into the Ka expression
Substitute the equilibrium concentrations into the Ka expression:
[
5.6 \times 10^{-10} = \frac{x \cdot x}{0.250 – x}
]
Since Ka is very small, we can assume that (x) is much smaller than 0.250, so we approximate (0.250 – x \approx 0.250). Thus, the equation becomes:
[
5.6 \times 10^{-10} = \frac{x^2}{0.250}
]
Solve for (x):
[
x^2 = (5.6 \times 10^{-10}) \times 0.250
]
[
x^2 = 1.4 \times 10^{-10}
]
[
x = \sqrt{1.4 \times 10^{-10}} = 1.18 \times 10^{-5} \, \text{M}
]
Step 5: Calculate pH
Since (x) represents the concentration of H⁺ ions, the pH is:
[
\text{pH} = -\log[H⁺] = -\log(1.18 \times 10^{-5}) \approx 4.93
]
Final Answer:
The pH of the 0.250 M NH₄Br solution is approximately 4.93.
Explanation:
Ammonium bromide (NH₄Br) dissociates in water to produce NH₄⁺, which is a weak acid. The Ka value for NH₄⁺ indicates that it dissociates to a small extent, producing H⁺ ions in the solution, which results in a slightly acidic pH. Using the provided Ka and the initial concentration, we used an ICE table and approximations to calculate the H⁺ concentration, and from that, the pH of the solution.