Question

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

Answer 1

Answer: The value of the equilibrium constant is 0.000023.

Explanation:

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

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

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

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

 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,

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

$K_c=\frac{(0.0015\times 0.0011}{(0.014\times 5.12)}$

$K_c=0.000023$

Thus the value of the equilibrium constant is 0.000023.