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Chapter 17: Problem 22

The deciding factor on why HF is a weak acid and not a strong acid like the other hydrogen halides is entropy. What occurs when HF dissociates in water as compared to the other hydrogen halides?

Short Answer

Expert verified
In summary, HF is a weak acid because its strong H-F bond and high electronegativity of fluorine make dissociation into ions difficult, limiting the increase in entropy that would favor dissociation. In contrast, other hydrogen halides have weaker H-X bonds and larger halide ions, making their dissociation more favorable in terms of increasing entropy, leading to higher acid strengths.

Step by step solution

01

Understand the terms weak and strong acids

First, let's clarify the meaning of weak and strong acids. A strong acid is one that completely ionizes in a solution, donating all of its protons (H+) to the solvent. In contrast, a weak acid only partially ionizes, donating only a small fraction of its protons. This means that strong acids have a higher concentration of H+ ions in the solution, leading to a lower pH.
02

Write the dissociation reactions for HF and other hydrogen halides

The general dissociation reaction for a hydrogen halide (HX) can be written as: \[ \mathrm{HX}\, (aq) \rightleftharpoons \mathrm{H^+} \, (aq) + \mathrm{X^-} \, (aq) \] Now, let's write the dissociation reactions for HF and another hydrogen halide (HCl, for example) in water: \[ \mathrm{HF}\, (aq) \rightleftharpoons \mathrm{H^+} \, (aq) + \mathrm{F^-} \, (aq) \] \[ \mathrm{HCl}\, (aq) \rightleftharpoons \mathrm{H^+} \, (aq) + \mathrm{Cl^-} \, (aq) \]
03

Analyze the bonds formed in the dissociation

When HF dissociates in water, it forms a hydrogen ion (H+) and a fluoride ion (F-). However, the H-F bond is highly polar, as fluorine is highly electronegative. The H-F bond is also relatively strong due to the small size of F-, compared to Cl- and other halide ions. Because of this strong bond, when the HF molecule splits into ions, the fluoride ion strongly attracts the proton, so less dissociation occurs in HF than in the other hydrogen halides. It makes HF a weak acid, despite its high polarity and ability to donate protons.
04

Understand the role of entropy

Entropy is a measure of the degree of disorder or randomness in a system. The dissociation of an acid into its ions increases the entropy of the system. However, the strong H-F bond makes it more difficult for the system to reach the higher entropy state associated with the ionized form. Comparing HF to other hydrogen halides, they have weaker H-X bonds and larger halide ions, making their dissociation more favorable in terms of increasing the entropy of the system.
05

Conclusion

In conclusion, HF is a weak acid compared to other hydrogen halides due to its strong polar H-F bond and high electronegativity of fluorine. These factors make it difficult for the HF molecule to fully dissociate into ions, limiting the increase in entropy that would be more favorable for dissociation. This is in contrast with other hydrogen halides, whose H-X bonds are weaker and more easily dissociated, leading to higher acid strengths.

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Most popular questions from this chapter

At 1 atm, liquid water is heated above \(100^{\circ} \mathrm{C}\) . For this process, which of the following choices (i-iv) is correct for \(\Delta S_{\text { surr }} ? \Delta S ?\) \(\Delta S_{\text { univ }} ?\) Explain each answer. i. greater than zero ii. less than zero iii. equal to zero iv. cannot be determined

Given the values of \(\Delta H\) and \(\Delta S,\) which of the following changes will be spontaneous at constant \(T\) and \(P ?\) a. \(\Delta H=+25 \mathrm{kJ}, \Delta S=+5.0 \mathrm{J} / \mathrm{K}, T=300 . \mathrm{K}\) b. \(\Delta H=+25 \mathrm{kJ}, \Delta S=+100 . \mathrm{J} / \mathrm{K}, T=300 . \mathrm{K}\) c. \(\Delta H=-10 . \mathrm{kJ}, \Delta S=+5.0 \mathrm{J} / \mathrm{K}, T=298 \mathrm{K}\) d. \(\Delta H=-10 . \mathrm{kJ}, \Delta S=-40 . \mathrm{J} / \mathrm{K}, T=200 . \mathrm{K}\)

Calculate the entropy change for the vaporization of liquid methane and liquid hexane using the following data. $$\begin{array}{lll} {\text { }} & {\text { Boiling Point (1 atm)}} & { \Delta H_{\mathrm{vap}} } \\ \hline {\text { Methane }} & \quad\quad\quad {112 \mathrm{K}} & {8.20 \mathrm{kJ} / \mathrm{mol}} \\ {\text { Hexane }} & \quad\quad\quad {342 \mathrm{K}} & {28.9 \mathrm{kJ} / \mathrm{mol}}\end{array}$$ Compare the molar volume of gaseous methane at 112 \(\mathrm{K}\) with that of gaseous hexane at 342 \(\mathrm{K}\) . How do the differences in molar volume affect the values of \(\Delta S_{\mathrm{vap}}\) for these liquids?

Some water is placed in a coffee-cup calorimeter. When 1.0 \(\mathrm{g}\) of an ionic solid is added, the temperature of the solution increases from \(21.5^{\circ} \mathrm{C}\) to \(24.2^{\circ} \mathrm{C}\) as the solid dissolves. For the dissolving process, what are the signs for \(\Delta S_{\mathrm{sys}}, \Delta S_{\mathrm{surt}},\) and \(\Delta S_{\mathrm{univ}} ?\)

Consider two perfectly insulated vessels. Vessel 1 initially contains an ice cube at \(0^{\circ} \mathrm{C}\) and water at \(0^{\circ} \mathrm{C}\) . Vessel 2 initially contains an ice cube at \(0^{\circ} \mathrm{C}\) and a saltwater solution at \(0^{\circ} \mathrm{C}\) . Consider the process \(\mathrm{H}_{2} \mathrm{O}(s) \rightarrow \mathrm{H}_{2} \mathrm{O}(l)\) a. Determine the sign of \(\Delta S, \Delta S_{\text { sum }}\) and \(\Delta S_{\text { univ }}\) for the process in vessel 1 . b. Determine the sign of \(\Delta S, \Delta S_{\text { sum }},\) and \(\Delta S_{\text { univ }}\) for the process in vessel \(2 .\) (Hint: Think about the effect that a salt has on the freezing point of a solvent.)
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