Which of the following solutions has the highest \(\left[\mathrm{H}^{+}\right]:\) (a) \(0.10 \mathrm{M} \mathrm{HF},\) (b) \(0.10 \mathrm{M} \mathrm{HF}\) in \(0.10 \mathrm{M} \mathrm{NaF}\) (c) \(0.10 \mathrm{M}\) HF in \(0.10 \mathrm{M} \mathrm{SbF}_{5} ?\) (Hint: SbF\(_{5}\) reacts with \(\mathrm{F}^{-}\) to form the complex ion \(\mathrm{SbF}_{6}^{-}\).)

HF is a weak acid and dissociates in water to form \(\mathrm{H}^{+}\) and \(\mathrm{F}^{-}\) ions. Hence, a \(0.10 \) M solution of HF will have a certain \(\left[\mathrm{H}^{+}\right]\) concentration. In solution (b), NaF is added. As NaF is a salt of a strong base (NaOH) and a weak acid (HF), it acts as a weak base and reduces the hydrogen ion concentration by forming HF. Therefore, solution (b) would have a lower \(\left[\mathrm{H}^{+}\right]\) concentration. In solution (c), SbF5 (a Lewis acid) reacts with the fluoride ion to form \(\mathrm{SbF}_{6}^{-}\) and reduces the concentration of \(\mathrm{F}^{-}\) ions. As the negative feedback for HF dissociation is removed, more HF can dissociate, resulting in a higher \(\left[\mathrm{H}^{+}\right]\) concentration.

Based on the descriptions in step 1, the solution (a) will have a higher \(\left[\mathrm{H}^{+}\right]\) concentration than solution (b) because the presence of NaF in solution (b) decreases the \(\left[\mathrm{H}^{+}\right]\) concentration. On the other hand, solution (c) will have the highest \(\left[\mathrm{H}^{+}\right]\) concentration due to the presence of SbF5. This is because SbF5 reduces the concentration of \(\mathrm{F}^{-}\) ions, allowing more HF to dissociate.