Determine the pH of a .22 M NaF solution. The Ka of HF is 3.5 x 10^-5
The Correct Answer and Explanation is :
To determine the pH of a 0.22 M NaF solution, we need to consider the hydrolysis of fluoride ions (F⁻) in water. Sodium fluoride (NaF) is a salt that dissociates completely in water into Na⁺ and F⁻. Since Na⁺ is a spectator ion, the fluoride ion undergoes hydrolysis:
[
F^- + H_2O \rightleftharpoons HF + OH^-
]
Step 1: Determine the ( K_b ) of ( F^- )
The equilibrium constant for this reaction is the base dissociation constant ( K_b ), which is related to the acid dissociation constant ( K_a ) of HF by:
[
K_b = \frac{K_w}{K_a}
]
where:
- ( K_w = 1.0 \times 10^{-14} ) (autoionization constant of water),
- ( K_a = 3.5 \times 10^{-5} ) (given for HF).
[
K_b = \frac{1.0 \times 10^{-14}}{3.5 \times 10^{-5}}
]
[
K_b = 2.86 \times 10^{-10}
]
Step 2: Set Up the ICE Table for Hydrolysis
Let ( x ) be the concentration of OH⁻ produced:
[
K_b = \frac{x^2}{0.22 – x} \approx \frac{x^2}{0.22}
]
[
2.86 \times 10^{-10} = \frac{x^2}{0.22}
]
Solving for ( x ):
[
x = \sqrt{(2.86 \times 10^{-10}) \times 0.22}
]
[
x = \sqrt{6.292 \times 10^{-11}}
]
[
x = 7.94 \times 10^{-6} \text{ M} \quad (\text{[OH⁻]})
]
Step 3: Determine pOH and pH
[
\text{pOH} = -\log(7.94 \times 10^{-6})
]
[
\text{pOH} \approx 5.10
]
[
\text{pH} = 14 – 5.10
]
[
\text{pH} \approx 8.90
]
Final Answer:
The pH of a 0.22 M NaF solution is 8.90.
Explanation:
Sodium fluoride is a basic salt because fluoride is the conjugate base of HF. When NaF dissolves, F⁻ hydrolyzes in water to produce OH⁻, making the solution basic. We determined the pH by calculating the ( K_b ) using the relationship with ( K_a ), solving for hydroxide concentration, and converting to pH. The result confirms that NaF solutions are mildly basic.