$$\ce{HF(aq) + H2O(l) <--> H3O+(aq) + F-(aq)}$$

The dissociation of the weak acid $\ce{HF}$ in water is represented by the equation above. Adding a 1.0 mL sample of which of the following would increase the percent ionization of HF(aq) in 10 mL of a solution of 1.0 $M  \ce{HF}$ ?

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$\ce{KF}$ will dissociate into $\ce{K+}$ and $\ce{F-}$. Increasing $\ce{F-}$ will shift the equilibrium position to the left.

$\ce{H2SO4}$ is an acid and will increase the $\ce{H3O+}$ concentration, shifting the equilibrium position to the left.

$\ce{HF}$ will dissociate into $\ce{H+}$ and $\ce{F-}$. Increasing $\ce{F-}$ will shift the equilibrium position to the left. As an acid, $\ce{HF}$ will also increase the hydronium concentration, shifting the equilibrium position to the left.

Since water is a liquid, it is not counted as a reactant when using Le Chatelier's principle. However, adding water does decrease the concentration of all species in solution. Since the concentration of products decreases faster than the reactants, the equilibrium position will shift to the right.