What is the pH of a 0.1 M NaF solution? Ionization constant for HF, Ka= 7 x 10
2.1
5.9
8.1
9.1
The Correct Answer and Explanation is :
To calculate the pH of a 0.1 M NaF solution, we need to consider the dissociation of NaF in water and how the fluoride ion (F⁻) behaves in solution.
- Dissociation of NaF: Sodium fluoride (NaF) dissociates completely in water:
[
\text{NaF} \rightarrow \text{Na}^+ + \text{F}^-
]
So, the concentration of fluoride ions (F⁻) in the solution is 0.1 M.
- Behavior of fluoride ions in water: Fluoride ions act as a base and can react with water to produce hydroxide ions (OH⁻) according to the following reaction:
[
\text{F}^- + \text{H}_2\text{O} \rightleftharpoons \text{HF} + \text{OH}^-
]
- Equilibrium expression: The equilibrium constant for this reaction, ( K_b ), can be calculated using the relationship between ( K_b ) and ( K_a ) (the ionization constant for HF):
[
K_w = K_a \times K_b
]
where ( K_w ) is the ionization constant for water (( K_w = 1 \times 10^{-14} )) and ( K_a ) is given as ( 7 \times 10^{-4} ) for HF.
We can solve for ( K_b ):
[
K_b = \frac{K_w}{K_a} = \frac{1 \times 10^{-14}}{7 \times 10^{-4}} = 1.43 \times 10^{-11}
]
- Calculating OH⁻ concentration: The equilibrium expression for the dissociation of F⁻ in water is:
[
K_b = \frac{[\text{HF}][\text{OH}^-]}{[\text{F}^-]}
]
Let ( x ) represent the concentration of OH⁻ produced at equilibrium. The equilibrium concentrations would be:
- ([F^-] = 0.1 – x)
- ([HF] = x)
- ([OH^-] = x)
Substituting into the equilibrium expression:
[
1.43 \times 10^{-11} = \frac{x^2}{0.1}
]
Solving for ( x ) (concentration of OH⁻):
[
x^2 = (1.43 \times 10^{-11}) \times 0.1 = 1.43 \times 10^{-12}
]
[
x = \sqrt{1.43 \times 10^{-12}} = 1.19 \times 10^{-6} \, \text{M}
]
This is the concentration of OH⁻.
- Calculating pOH: The pOH is given by:
[
\text{pOH} = -\log[\text{OH}^-] = -\log(1.19 \times 10^{-6}) = 5.92
]
- Calculating pH: The pH is related to the pOH by the equation:
[
\text{pH} + \text{pOH} = 14
]
Thus,
[
\text{pH} = 14 – 5.92 = 8.08
]
Therefore, the pH of the 0.1 M NaF solution is approximately 8.1.
Answer: 8.1.
I will now generate a visual explanation of the calculation.
Here is a detailed diagram illustrating the process of calculating the pH of a 0.1 M NaF solution. It covers each step from dissociation to the final pH calculation. Let me know if you need further clarification!
