What is the pH of 0.10 M NaF(aq). The Ka of HF is 6.8 x 10-4
A. 2.08
B. 8.08
C. 5.92
D. 14.92
The Correct Answer and Explanation is:
The pH of a 0.10 M NaF solution can be determined by considering the behavior of NaF in water. Sodium fluoride (NaF) dissociates completely into Na⁺ and F⁻ ions. The fluoride ion (F⁻) is the conjugate base of hydrofluoric acid (HF). To calculate the pH, we need to determine the concentration of OH⁻ ions in the solution due to the hydrolysis of F⁻.
Step 1: Write the hydrolysis reaction for F⁻
The fluoride ion (F⁻) reacts with water in the following equilibrium reaction:
[
F⁻(aq) + H_2O(l) \rightleftharpoons HF(aq) + OH⁻(aq)
]
The fluoride ion accepts a proton (H⁺) from water to form HF and releases hydroxide ions (OH⁻), which makes the solution basic.
Step 2: Set up an expression for the equilibrium constant (Kb)
To calculate the pH, we first need to calculate the base dissociation constant (Kb) for F⁻. We can use the relation between Ka and Kb for conjugate acid-base pairs:
[
K_b = \frac{K_w}{K_a}
]
where (K_w) is the ion product of water, (1.0 \times 10^{-14}), and (K_a) is the acid dissociation constant for HF, which is given as (6.8 \times 10^{-4}).
[
K_b = \frac{1.0 \times 10^{-14}}{6.8 \times 10^{-4}} = 1.47 \times 10^{-11}
]
Step 3: Set up the ICE table and solve for [OH⁻]
Using the initial concentration of F⁻, which is 0.10 M, we can set up an ICE (Initial, Change, Equilibrium) table for the hydrolysis reaction:
| Species | Initial concentration | Change in concentration | Equilibrium concentration |
|---|---|---|---|
| F⁻ | 0.10 M | -x | 0.10 – x |
| HF | 0 | +x | x |
| OH⁻ | 0 | +x | x |
Using the expression for Kb:
[
K_b = \frac{[HF][OH⁻]}{[F⁻]} = \frac{x^2}{0.10 – x}
]
Since (K_b) is very small, we can assume that (x) is much smaller than 0.10 M, so (0.10 – x \approx 0.10). Thus, the equation simplifies to:
[
1.47 \times 10^{-11} = \frac{x^2}{0.10}
]
Solving for (x):
[
x^2 = (1.47 \times 10^{-11}) \times 0.10 = 1.47 \times 10^{-12}
]
[
x = \sqrt{1.47 \times 10^{-12}} = 1.21 \times 10^{-6} \text{ M}
]
This is the concentration of OH⁻.
Step 4: Calculate the pOH and pH
The pOH is related to the concentration of OH⁻:
[
pOH = -\log[OH⁻] = -\log(1.21 \times 10^{-6}) = 5.92
]
Finally, the pH is calculated using the relation:
[
pH = 14 – pOH = 14 – 5.92 = 8.08
]
Final Answer:
The pH of the 0.10 M NaF solution is 8.08.
Thus, the correct answer is B. 8.08.