Question

The weak ionization constant (Ka)

for HF is equal to:
[H3O][F]
[HFJ[H20]
[HF]
[F-][H30+]
Please help

Answer 1

`\displaystyle K_a = \frac{[\text{H$_3$O}^+][\text{F}^-]}{[\text{HF}]}`

Step-by-step explanation:

Write the reaction of HF (a weak acid) with water:

`\displaystyle \text{HF}_\text{(aq)} + \text{H$_2$O}_\text{($\ell$)} \rightleftharpoons \text{H$_3$O}^+_\text{(aq)} + \text{F}^-_\text{(aq)}`

Recall that the equilibrium constant expression for an equilibrium reaction is the product of the concentration of the products over the product of the concentration of the reactants raised to their respective coefficients.

The equilibrium constant expression for this equation will hence be:

`\displaystyle K_a = \frac{[\text{H$_3$O}^+][\text{F}^-]}{[\text{HF}]}`

Note that pure solids and liquids are not included in the equilibrium expression.

In conclusion:

`\displaystyle K_a = \frac{[\text{H$_3$O}^+][\text{F}^-]}{[\text{HF}]}`