Question

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.

Answer 1

Answer: The value of the equilibrium constant is 0.000023.

Step-by-step explanation:

Equilibrium concentration of
`HF` =
`(1.62g)/(20.01g/mol* 5.6L)=0.014M`

Equilibrium concentration of
`H_2O` =
`(516g)/(18g/mol* 5.6L)=5.12M`

Equilibrium concentration of
`F^-` =
`(0.163g)/(18.9g/mol* 5.6L)=0.0015M`

Equilibrium concentration of
`H_3O^+` =
`(0.110g)/(18g/mol* 5.6L)=0.0011M`

The given balanced equilibrium reaction is,

`HF(aq)+H_2O(l)\rightleftharpoons F^-(aq)+H_3O^+(aq)`

The expression for equilibrium constant for this reaction will be,

`K_c=([F^-]* [H_3O^+])/([HF]* [H_2O])`

`K_c=((0.0015* 0.0011)/((0.014* 5.12))`

`K_c=0.000023`

Thus the value of the equilibrium constant is 0.000023.