Hydrofluoric acid and water react to form fluoride anion and hydronium cation, like this: HF (aq)+H₂O()-F (aq) + H₂O (aq) At a certain temperature, a chemist finds that a 6.5 L. reaction vessel containing an aqueous solution of hydrofluoric acid, water, fluoride anion, and hydronium cation at equilibrium has the following composition: amount. 1.64 g 639. B F 0.199 g H,O' 0.384 g Calculate the value of the equilibrium constant K, for this reaction. Round your answer to 2 significant digits. compound HF H₂O K-D 0.8 X

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**Chemical Equilibrium and Equilibrium Constant Calculation**

Hydrofluoric acid and water react to form fluoride anion and hydronium cation as shown in the chemical equation:

\[ \text{HF}_{(aq)} + \text{H}_2\text{O}_{(l)} \rightarrow \text{F}^-_{(aq)} + \text{H}_3\text{O}^+_{(aq)} \]

In a scenario at a certain temperature, a chemist analyzes a 6.5 L reaction vessel containing an aqueous solution of hydrofluoric acid, water, fluoride ion, and hydronium ion at equilibrium. The observed composition at equilibrium is as follows:

| Compound | Amount (grams) |
|----------|----------------|
| HF       | 1.64 g         |
| H₂O      | 639 g          |
| F⁻       | 0.199 g        |
| H₃O⁺     | 0.384 g        |

**Objective:**

Calculate the value of the equilibrium constant \( K_{\text{c}} \) for this reaction. Round your answer to 2 significant digits.

**Instructions:**

1. Convert the given masses into moles using the respective molar masses.
2. Calculate the molarity of each species by dividing the number of moles by the volume of the solution (6.5 L).
3. Use the equilibrium expression for the reaction to calculate \( K_{\text{c}} \).

**Equilibrium Expression:**

\[ K_{\text{c}} = \frac{[\text{F}^-][\text{H}_3\text{O}^+]}{[\text{HF}]} \]

(Note: The concentration of water is typically omitted in \( K_{\text{c}} \) expressions for aqueous reactions.)

This calculation enables a deeper understanding of chemical equilibria and the dynamics of reversible reactions in aqueous solutions.
Transcribed Image Text:**Chemical Equilibrium and Equilibrium Constant Calculation** Hydrofluoric acid and water react to form fluoride anion and hydronium cation as shown in the chemical equation: \[ \text{HF}_{(aq)} + \text{H}_2\text{O}_{(l)} \rightarrow \text{F}^-_{(aq)} + \text{H}_3\text{O}^+_{(aq)} \] In a scenario at a certain temperature, a chemist analyzes a 6.5 L reaction vessel containing an aqueous solution of hydrofluoric acid, water, fluoride ion, and hydronium ion at equilibrium. The observed composition at equilibrium is as follows: | Compound | Amount (grams) | |----------|----------------| | HF | 1.64 g | | H₂O | 639 g | | F⁻ | 0.199 g | | H₃O⁺ | 0.384 g | **Objective:** Calculate the value of the equilibrium constant \( K_{\text{c}} \) for this reaction. Round your answer to 2 significant digits. **Instructions:** 1. Convert the given masses into moles using the respective molar masses. 2. Calculate the molarity of each species by dividing the number of moles by the volume of the solution (6.5 L). 3. Use the equilibrium expression for the reaction to calculate \( K_{\text{c}} \). **Equilibrium Expression:** \[ K_{\text{c}} = \frac{[\text{F}^-][\text{H}_3\text{O}^+]}{[\text{HF}]} \] (Note: The concentration of water is typically omitted in \( K_{\text{c}} \) expressions for aqueous reactions.) This calculation enables a deeper understanding of chemical equilibria and the dynamics of reversible reactions in aqueous solutions.
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