Question

The ionisation constant of the weak acid HF whose concentration is 0.1 M is \( 3.5 \times 10^{-4} \). The equilibrium constant value for the reaction
\[ \text{F}^- + \text{H}_2\text{O} \leftrightarrow \text{HF} + \text{OH}^- \]
is ............ and the pH of the aqueous solution of the weak acid is ...............

• A. \( K_{\text{eq}} = 4.4 \times 10^{-11} \quad \text{pH} = 4.25 \)
• B. \( K_{\text{eq}} = 3.58 \times 10^{-11} \quad \text{pH} = 4.19 \)
• C. \( K_{\text{eq}} = 2.86 \times 10^{-11} \quad \text{pH} = 3.23 \)
• D. \( K_{\text{eq}} = 3.92 \times 10^{-10} \quad \text{pH} = 4.07 \)

Hint:

When dealing with weak acid dissociation and equilibrium constants, remember that the dissociation of water and the conjugate base affect the equilibrium. Also, the pH can be calculated from the \( [\text{H}^+] \) concentration using \( \text{pH} = -\log [\text{H}^+] \).

Answer 1

The Correct Option is C

Step 1: Write the equilibrium expression for the ionisation constant of HF.
The ionisation constant for the weak acid HF is given as:
\[ K_a = \frac{[\text{H}^+][\text{F}^-]}{[\text{HF}]} = 3.5 \times 10^{-4} \]
We can use this to find the concentration of \( \text{H}^+ \) ions in the solution.

Step 2: Find the equilibrium concentration of \( \text{OH}^- \) from the given reaction.
For the reaction:
\[ \text{F}^- + \text{H}_2\text{O} \leftrightarrow \text{HF} + \text{OH}^- \]
the equilibrium constant \( K_{\text{eq}} \) is related to the ionisation constant of HF and the dissociation of water.
The expression for \( K_{\text{eq}} \) is:
\[ K_{\text{eq}} = \frac{[\text{HF}][\text{OH}^-]}{[\text{F}^-][\text{H}_2\text{O}]} \]
Since water is in excess, we can simplify this to:
\[ K_{\text{eq}} = \frac{[\text{HF}][\text{OH}^-]}{[\text{F}^-]} \] We can now calculate the equilibrium constant value \( K_{\text{eq}} \).

Step 3: Find the pH of the solution.
The pH of the solution is given by:
\[ \text{pH} = -\log [\text{H}^+] \]
We can calculate \( [\text{H}^+] \) using the ionisation constant and the concentration of HF.

Step 4: Conclusion.
After solving the equations, we find that the equilibrium constant \( K_{\text{eq}} = 2.86 \times 10^{-11} \) and the pH of the solution is 3.23. Therefore, the correct answer is option (C).