Respuesta :
The question is incomplete , complete question is:
Hydrofluoric acid and water react to form fluoride anion and hydronium cation, like this:
[tex]HF(aq)+H_2O(l)\rightarrow F^-(aq)+H_3O^+(aq) [/tex]
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
[tex]HF[/tex] = 1.62 g
[tex]H_2O[/tex] = 516 g
[tex]F^-[/tex] = 0.163 g
[tex]H_3O^+[/tex] = 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 [tex]1.095\times 10^{-4}[/tex].
Explanation:
[tex]F^-(aq)+H_3O^+(aq)\rightleftharpoons HF(aq)+H_2O(l)[/tex]
[tex]Concentration =\frac{n}{V}[/tex]
Where :
n = moles of compound
V = Volume of the solution L
Moles of HF = [tex]\frac{1.62 g}{20 g/mol}=0.081 mol[/tex]
Volume of the solution in the vessel = V = 5.6 L
[tex][HF]=\frac{0.081 mol}{5.6 L}=0.01446 M[/tex]
Moles of [tex]F^-\frac{0.163 g}{19g/mol}=0.008579 mol[/tex]
Volume of the solution in the vessel = V = 5.6 L
[tex][F^-]=\frac{0.008579 mol}{5.6 L}=0.001532 M[/tex]
Moles of [tex]H_3O^+=\frac{0.110 g}{19 g/mol}=0.05789 mol[/tex]
Volume of the solution in the vessel = V = 5.6 L
[tex][H_3O^+]=\frac{0.05789 mol}{5.6 L}=0.001034 M[/tex]
An equilibrium expression is given as;
[tex]K_c=\frac{[F^-][H_3O^+]}{[HF]}[/tex]
[tex]=\frac{0.001532 M\times 0.001034 M}{0.01446 M}[/tex]
[tex]K_c=0.0001095=1.095\times 10^{-4}[/tex]
The value of equilibrium constant of the reaction is [tex]1.095\times 10^{-4}[/tex].
