Working with Aqueous Solutions of HF and NaF
a. What is the pH of a 0.35 M hydrofluoric acid (HF) solution? Ka 6.8x 104 [1.81]
b. What is the pH of a solution containing 0.35 M HF and 0.28 M NaF? [3.07]
The Correct Answer and Explanation is :
Let’s go step by step for both calculations.
(a) Finding the pH of a 0.35 M HF Solution
Hydrofluoric acid (HF) is a weak acid that partially ionizes in water:
[
HF \rightleftharpoons H^+ + F^-
]
Given:
- Initial concentration of HF, ( [HF] = 0.35 ) M
- Acid dissociation constant, ( K_a = 6.8 \times 10^{-4} )
Using the ICE (Initial, Change, Equilibrium) table and the expression for ( K_a ):
[
K_a = \frac{[H^+][F^-]}{[HF]}
]
Let ( x ) be the concentration of ( H^+ ) that dissociates. At equilibrium:
[
K_a = \frac{x^2}{0.35 – x}
]
Since ( K_a ) is small, we approximate ( 0.35 – x \approx 0.35 ), so:
[
6.8 \times 10^{-4} = \frac{x^2}{0.35}
]
Solving for ( x ):
[
x^2 = (6.8 \times 10^{-4}) \times (0.35)
]
[
x^2 = 2.38 \times 10^{-4}
]
[
x = \sqrt{2.38 \times 10^{-4}} = 0.0154
]
Thus,
[
\text{pH} = -\log(0.0154) = 1.81
]
(b) Finding the pH of a Buffer Solution (0.35 M HF and 0.28 M NaF)
This is a buffer solution because it contains both a weak acid (HF) and its conjugate base (( F^- )) from NaF. We use the Henderson-Hasselbalch equation:
[
\text{pH} = pK_a + \log \frac{[A^-]}{[HA]}
]
First, calculate ( pK_a ):
[
pK_a = -\log(6.8 \times 10^{-4}) = 3.17
]
Now, plug in the concentrations:
[
\text{pH} = 3.17 + \log \frac{0.28}{0.35}
]
[
\text{pH} = 3.17 + \log(0.8)
]
[
\text{pH} = 3.17 + (-0.10) = 3.07
]
Explanation (300 Words)
Hydrofluoric acid (HF) is a weak acid because it does not completely dissociate in water. The dissociation is controlled by its acid dissociation constant (( K_a )), which is ( 6.8 \times 10^{-4} ). To determine the pH of a pure HF solution, we use an ICE table to set up an equilibrium equation based on ( K_a ). Since ( K_a ) is relatively small, we approximate the equilibrium expression, solving for ( x ), the ( H^+ ) concentration, to find the pH of 1.81.
When sodium fluoride (NaF) is added, it dissociates completely into ( Na^+ ) and ( F^- ). The presence of fluoride ions suppresses the dissociation of HF due to Le Chatelier’s principle. This creates a buffer system, where HF (weak acid) and ( F^- ) (conjugate base) resist changes in pH.
To calculate the pH of the buffer, we apply the Henderson-Hasselbalch equation, which is derived from the acid dissociation expression. The equation relates pH to the ( pK_a ) and the ratio of the conjugate base (( F^- )) to the weak acid (HF). Substituting given concentrations, we find the pH to be 3.07.
Buffers are crucial in biological and chemical systems as they stabilize pH. This HF/NaF buffer system, for example, is used in fluoride treatments and industrial applications.