Which of the following could be added to an aqueous solution of weak acid HF to increase the percent dissociation? (A) \(\operatorname{NaF}(s)\) (B) \(\mathrm{H}_{2} \mathrm{O}(l)\) (C) \(\mathrm{NaOH}(\mathrm{s})\) (D) \(\mathrm{NH}_{3}(a q)\)

HF is a weak acid which partially dissociates in water giving the following equilibrium reaction: \(HF(aq) \rightleftharpoons H^+(aq) + F^-(aq)\). In this case, anything that would encourage the dissociation (forward reaction) would increase the percent dissociation.

(A) \(\operatorname{NaF}(s)\): This is a salt of the weak acid. When in solution, it dissociates completely to give \(\operatorname{Na}^+\) and \(\operatorname{F}^-\) ions. According to Le Chatelier's principle, adding more \(\operatorname{F}^-\) ions will shift the equilibrium to the left, hindering further dissociation of HF. (B) \(\mathrm{H}_{2} \mathrm{O}(l)\): While water is the solvent in which HF is dissolved, adding more water does not necessarily change the percent dissociation as it doesn't change the equilibrium directly. (C) \(\mathrm{NaOH}(\mathrm{s})\): This is a strong base and will react with HF to consume the \(\mathrm{H}^+\) ions (NaOH + HF → NaF + H2O), which would shift the equilibrium to the right causing more HF to dissociate. (D) \(\mathrm{NH}_{3}(a q)\): Ammonia is a weak base and will react with \(\mathrm{H}^+\) ions to a lesser extent than NaOH, but it will still cause a shift in the equilibrium towards right and increase the dissociation of HF. However, the influence of NH3 is considerably less than NaOH.

Considering all the analysis, NaOH(s) would cause the greatest increase in the percent dissociation of HF, because it is a strong base that would shift the equilibrium to the right, causing more dissociation of HF. Despite NH3 also causing a similar effect, it won't be as impactful as NaOH due to its weak base nature.