lithium fluoride, LIF, is a salt formed from the neutralization of the weak acid hydrofluoric acid, HF, with the strong base lithium hydroxide. Given that the value of KA for hydrofluoric acid is 3.5 * 10^-4 , what is the pH of a 0.187 m solution of lithium fluoride at 23 degrees Celsius? your answer should have three significant figures
The Correct Answer and Explanation is :
To calculate the pH of a 0.187 M solution of lithium fluoride (LiF), we need to consider the dissociation of the salt in water. Lithium fluoride is a salt formed by the neutralization of the weak acid hydrofluoric acid (HF) with the strong base lithium hydroxide (LiOH).
Step 1: Understanding the Nature of Lithium Fluoride
When lithium fluoride dissolves in water, it dissociates as follows:
[
LiF (s) \rightarrow Li^+ (aq) + F^- (aq)
]
The fluoride ion (F⁻) is the conjugate base of hydrofluoric acid (HF). Since HF is a weak acid (with a given acid dissociation constant, ( K_a ) of ( 3.5 \times 10^{-4} )), the fluoride ion will undergo hydrolysis, reacting with water to form HF and OH⁻:
[
F^- (aq) + H_2O (l) \rightleftharpoons HF (aq) + OH^- (aq)
]
Step 2: Setting up the Equilibrium Expression
The equilibrium constant for the hydrolysis reaction can be derived from the ( K_a ) of HF and the ( K_b ) of F⁻. The relation between ( K_a ) and ( K_b ) for a conjugate acid-base pair is given by:
[
K_b = \frac{K_w}{K_a}
]
Where ( K_w ) is the ion-product constant for water at 23°C, which is ( 1.0 \times 10^{-14} ). So, the ( K_b ) for F⁻ is:
[
K_b = \frac{1.0 \times 10^{-14}}{3.5 \times 10^{-4}} = 2.86 \times 10^{-11}
]
Step 3: Setting up the ICE Table
We now set up an ICE (Initial, Change, Equilibrium) table for the hydrolysis of fluoride:
| Species | Initial Concentration | Change in Concentration | Equilibrium Concentration |
|---|---|---|---|
| F⁻ | 0.187 M | -x | 0.187 – x |
| H₂O | (constant) | – | – |
| HF | 0 | +x | x |
| OH⁻ | 0 | +x | x |
Step 4: Solving for ( x )
Using the equilibrium expression for ( K_b ):
[
K_b = \frac{[HF][OH^-]}{[F^-]}
]
Substituting the equilibrium concentrations:
[
2.86 \times 10^{-11} = \frac{x^2}{0.187 – x}
]
Since ( K_b ) is very small, we can approximate ( 0.187 – x \approx 0.187 ). Thus:
[
2.86 \times 10^{-11} = \frac{x^2}{0.187}
]
Solving for ( x ):
[
x^2 = (2.86 \times 10^{-11}) \times 0.187
]
[
x^2 = 5.34 \times 10^{-12}
]
[
x = \sqrt{5.34 \times 10^{-12}} = 7.31 \times 10^{-6} \, \text{M}
]
Step 5: Finding the pOH and pH
The concentration of OH⁻ is ( 7.31 \times 10^{-6} ) M. To find the pOH:
[
\text{pOH} = -\log[OH^-] = -\log(7.31 \times 10^{-6}) = 5.137
]
Finally, to find the pH, we use the relation:
[
\text{pH} = 14 – \text{pOH} = 14 – 5.137 = 8.863
]
Final Answer:
The pH of the 0.187 M solution of lithium fluoride is 8.86.
Explanation:
Lithium fluoride dissociates into Li⁺ and F⁻ ions. The fluoride ion, as the conjugate base of the weak acid HF, reacts with water to form hydroxide ions, leading to an increase in the pH of the solution. The equilibrium calculation involved finding the concentration of OH⁻ formed by this hydrolysis reaction, which allowed us to compute the pOH and, subsequently, the pH of the solution.