Hydrofluoric acid and water react to form fluoride anion and hydronium cation, like this:
HF(aq) + H2O(I) rightarrow F-(aq) + H3O+(aq)
At a certain temperature, a chemist finds that a 5.6 L reaction vessel containing an aqueous solution of hydrofluoric acid, water, fluoride anion, and hydronium cation at equilibrium has the following composition:
Compound Amount
HF 1.62 g
H2O 516 g
F- 0.163 g
H3O+ 0.110 g
Calculate the value of the equilibrium constant for this reaction.

Question

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Respuesta :

kobenhavn kobenhavn · Answer 1 · 2021-04-02T05:08:28+02:00

Answer: The value of the equilibrium constant is 0.000023.

Explanation:

Equilibrium concentration of [tex]HF[/tex] = [tex]\frac{1.62g}{20.01g/mol\times 5.6L}=0.014M[/tex]

Equilibrium concentration of [tex]H_2O[/tex] = [tex]\frac{516g}{18g/mol\times 5.6L}=5.12M[/tex]

Equilibrium concentration of [tex]F^-[/tex] = [tex]\frac{0.163g}{18.9g/mol\times 5.6L}=0.0015M[/tex]

Equilibrium concentration of [tex]H_3O^+[/tex] = [tex]\frac{0.110g}{18g/mol\times 5.6L}=0.0011M[/tex]

The given balanced equilibrium reaction is,

[tex]HF(aq)+H_2O(l)\rightleftharpoons F^-(aq)+H_3O^+(aq)[/tex]

The expression for equilibrium constant for this reaction will be,

[tex]K_c=\frac{[F^-]\times [H_3O^+]}{[HF]\times [H_2O]}[/tex]

[tex]K_c=\frac{(0.0015\times 0.0011}{(0.014\times 5.12)}[/tex]

[tex]K_c=0.000023[/tex]

Thus the value of the equilibrium constant is 0.000023.