Chapter 17: Problem 104
Which of these solutions has the highest \(\left[\mathrm{H}^{+}\right]\) : (a) \(0.10 \mathrm{M} \mathrm{HF},\) (b) \(0.10 \mathrm{M} \mathrm{HF}\) in \(0.10 \mathrm{M} \mathrm{NaF}\), or (c) \(0.10 \mathrm{M} \mathrm{HF}\) in \(0.10 \mathrm{M} \mathrm{SbF}_{5}\) ? (Hint: \(\mathrm{SbF}_{5}\) reacts with \(\mathrm{F}^{-}\) to form the complex ion \(\mathrm{SbF}_{6}^{-}\).)
Short Answer
Step by step solution
Understanding the Acid Dissociation
Analyzing Solution (a): Pure 0.10 M HF
Analyzing Solution (b): 0.10 M HF in 0.10 M NaF
Analyzing Solution (c): 0.10 M HF in 0.10 M SbF5
Comparing \( [\text{H}^+] \) in Each Solution
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Common Ion Effect
When NaF dissolves, it readily breaks up into \( \text{Na}^+ \) and \( \text{F}^- \). This influx of \( \text{F}^- \) from NaF affects the equilibrium of HF dissociation. The presence of a common ion (\( \text{F}^- \) here) suppresses the dissociation of the weak acid, HF, in a solution.
- Due to this increase in \( \text{F}^- \) concentration from NaF, the dissociation equilibrium of HF shifts, reducing its breakdown into \( \text{H}^+ \) and \( \text{F}^- \) ions.
- As a result, fewer hydrogen ions are produced, hence decreasing the solution's acidity. \( \text{H}^+ \)
Le Chatelier's Principle
Le Chatelier's Principle suggests that when a change is imposed on a system at equilibrium, the system will adjust to counteract the change and restore a new equilibrium.
- For example, in our exercise, when \( \text{F}^- \) ions from NaF are added to a solution of HF, it increases the concentration of fluoride ions, which is a product of HF dissociation.
- According to Le Chatelier's Principle, the equilibrium will shift towards the reactants to reduce the effect of this increase in \( \text{F}^- \), essentially inhibiting further dissociation of HF.
Equilibrium Concentration
In simple terms, the equilibrium concentration represents the concentrations of different ions in the reversible reaction at equilibrium.
- For HF in pure water, the equilibrium concentration of \( \text{H}^+ \) is determined by how much HF dissociates.
- In a mixed solution with NaF, the equilibrium concentrations of ions change because of the common ion effect, which increases \( \text{F}^- \) and decreases \( \text{H}^+ \) due to suppressed dissociation.
Acid Dissociation Constant
The equation for a weak acid such as HF is represented as: \[\text{HF} \rightleftharpoons \text{H}^+ + \text{F}^-\] \( K_a \) is determined by the ratio:\[K_a = \frac{[\text{H}^+][\text{F}^-]}{[\text{HF}]}\]
- For HF, \( K_a \) is around \( 6.8 \times 10^{-4} \), indicating that only a small portion dissociates in water.
- This low \( K_a \) value shows that HF is a weak acid, only partially contributing to \( \text{H}^+ \) in solution.