HF(aq) + H2O(l) ⇄ H3O+(aq) + F−(aq) The dissociation of the weak acid HF in water is represented by — Acids and Bases Chemistry Question

STEPS:

1. Define percent ionization: The percent ionization of a weak acid like hydrofluoric acid ($\text{HF}$) represents the percentage of the initial acid molecules that dissociate into ions at equilibrium:
$$\text{Percent Ionization} = \frac{[\text{F}^-]_{\text{eq}}}{[\text{HF}]_{\text{initial}}} \times 100\%$$
2. Understand the effect of dilution on concentration: When distilled water is added to the solution, the volume of the mixture increases, which immediately dilutes (decreases) the concentrations of all dissolved species ($\text{HF}$, $\text{H}_3\text{O}^+$, and $\text{F}^-$).
3. Analyze the reaction quotient ($Q$) after dilution: The equilibrium constant expression for the dissociation is defined as:
$$K_a = \frac{[\text{H}_3\text{O}^+][\text{F}^-]}{[\text{HF}]}$$
Upon dilution by a volume factor $V$, the concentration of each aqueous species is divided by $V$. The reaction quotient immediately after dilution becomes:
$$Q = \frac{\left(\frac{[\text{H}_3\text{O}^+]}{V}\right)\left(\frac{[\text{F}^-]}{V}\right)}{\left(\frac{[\text{HF}]}{V}\right)} = \frac{1}{V} \left( \frac{[\text{H}_3\text{O}^+][\text{F}^-]}{[\text{HF}]} \right) = \frac{1}{V} K_a$$
Since the volume increased ($V > 1$), the reaction quotient is now smaller than the equilibrium constant ($Q < K_a$).
4. Determine the direction of the equilibrium shift: According to Le Chatelier's Principle, the system must shift to counteract this dilution. Because there are more aqueous particles on the product side (two ions) than on the reactant side (one molecule), shifting in the forward direction (to the right) generates more particles to partially offset the decrease in concentration.
5. Relate the shift to percent ionization: While the final equilibrium *concentrations* of all species will be slightly lower than they were initially due to the larger volume, the forward shift means a greater net number of moles of HF dissociate compared to the original solution. Because more moles of $\text{HF}$ have ionized out of the same original quantity, the percent ionization increases. Therefore, adding distilled water (Option D) increases the percent ionization of the weak acid.

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WHY_OTHERS_WRONG:

* Option A is incorrect: Adding $1.0\text{ M KF}$ introduces fluoride ions ($\text{F}^-$), which is a product of the dissociation reaction. This increases the product concentration, triggering the common-ion effect. The equilibrium shifts to the left to consume the excess product, reducing the percent ionization of $\text{HF}$.
* Option B is incorrect: Adding $1.0\text{ M }\text{H}_2\text{SO}_4$ introduces a high concentration of hydronium ions ($\text{H}_3\text{O}^+$) because sulfuric acid is a strong acid. This drastically increases the concentration of a product ion, driving the equilibrium to the left via the common-ion effect and severely reducing the percent ionization of $\text{HF}$.
* Option C is incorrect: Adding highly concentrated $10.0\text{ M }\text{HF}$ increases the overall concentration of the acid in the solution. For any weak acid, as the starting concentration of the acid increases, the percent ionization decreases. This is because the concentration of dissociated ions at equilibrium is proportional to the square root of the initial concentration ($\propto \sqrt{K_a \cdot C_0}$), meaning the ionized fraction ($\frac{\sqrt{K_a \cdot C_0}}{C_0} \propto \sqrt{\frac{K_a}{C_0}}$) gets smaller as $C_0$ increases.