Question

Hydrofluoric acid and water react to form fluoride anion and hydronium cation, like this: (aq)(l)(aq)(aq) At a certain temperature, a chemist finds that a reaction vessel containing an aqueous solution of hydrofluoric acid, water, fluoride anion, and hydronium cation at equilibrium has the following composition: compound amount Calculate the value of the equilibrium constant for this reaction. Round your answer to significant digits.

Answer 1

The question is incomplete , complete question is:

Hydrofluoric acid and water react to form fluoride anion and hydronium cation, like this:

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

At a certain temperature, a chemist finds that a reaction vessel 5.6 L 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

`H_2O` = 516 g

`F^-` = 0.163 g

`H_3O^+` = 0.110 g

Calculate the value of the equilibrium constant for this reaction. Round your answer to significant digits.

Answer:

The value of equilibrium constant of the reaction is
`1.095* 10^(-4)`.

Step-by-step explanation:

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

`Concentration =(n)/(V)`

Where :

n = moles of compound

V = Volume of the solution L

Moles of HF =
`(1.62 g)/(20 g/mol)=0.081 mol`

Volume of the solution in the vessel = V = 5.6 L

`[HF]=(0.081 mol)/(5.6 L)=0.01446 M`

Moles of
`F^-(0.163 g)/(19g/mol)=0.008579 mol`

Volume of the solution in the vessel = V = 5.6 L

`[F^-]=(0.008579 mol)/(5.6 L)=0.001532 M`

Moles of
`H_3O^+=(0.110 g)/(19 g/mol)=0.05789 mol`

Volume of the solution in the vessel = V = 5.6 L

`[H_3O^+]=(0.05789 mol)/(5.6 L)=0.001034 M`

An equilibrium expression is given as;

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

`=(0.001532 M* 0.001034 M)/(0.01446 M)`

`K_c=0.0001095=1.095* 10^(-4)`

The value of equilibrium constant of the reaction is
`1.095* 10^(-4)`.