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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

You remember that \(\Delta G^{\circ}\) is related to \(R T \ln (K)\) but cannot remember if it's \(R T \ln (K)\) or \(-R T \ln (K)\) . Realizing what \(\Delta G^{\circ}\) and K mean, how can you figure out the correct sign?

Impure nickel, refined by smelting sulfide ores in a blast furnace, can be converted into metal from 99.90% to 99.99% purity by the Mond process. The primary reaction involved in the Mond process is $$\mathrm{Ni}(s)+4 \mathrm{CO}(g) \rightleftharpoons \mathrm{Ni}(\mathrm{CO})_{4}(g)$$ a. Without referring to Appendix \(4,\) predict the sign of \(\Delta S^{\circ}\) for the above reaction. Explain. b. The spontaneity of the above reaction is temperature-dependent. Predict the sign of \(\Delta S_{\text { sum }}\) for this reaction. Explain. c. For \(\mathrm{Ni}(\mathrm{CO})_{4}(g), \Delta H_{\mathrm{f}}^{\circ}=-607 \mathrm{kJ} / \mathrm{mol}\) and \(S^{\circ}=417 \mathrm{J} / \mathrm{K} \cdot \mathrm{mol}\) at 298 \(\mathrm{K}\) . Using these values and data in Appendix 4 calculate \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) for the above reaction. d. Calculate the temperature at which \(\Delta G^{\circ}=0(K=1)\) for the above reaction, assuming that \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) do not depend on temperature. e. The first step of the Mond process involves equilibrating impure nickel with \(\mathrm{CO}(g)\) and \(\mathrm{Ni}(\mathrm{CO})_{4}(g)\) at about \(50^{\circ} \mathrm{C} .\) The purpose of this step is to convert as much nickel as possible into the gas phase. Calculate the equilibrium constant for the above reaction at \(50 .^{\circ} \mathrm{C}\) f. In the second step of the Mond process, the gaseous \(\mathrm{Ni}(\mathrm{CO})_{4}\) is isolated and heated to \(227^{\circ} \mathrm{C}\) . The purpose of this step is to deposit as much nickel as possible as pure solid (the reverse of the preceding reaction). Calculate the equilibrium constant for the preceding reaction at \(227^{\circ} \mathrm{C}\) . g. Why is temperature increased for the second step of the Mond process? h. The Mond process relies on the volatility of \(\mathrm{Ni}(\mathrm{CO})_{4}\) for its success. Only pressures and temperatures at which \(\mathrm{Ni}(\mathrm{CO})_{4}\) is a gas are useful. A recently developed variation of the Mond process carries out the first step at higher pressures and a temperature of \(152^{\circ} \mathrm{C}\) . Estimate the maximum pressure of \(\mathrm{Ni}(\mathrm{CO})_{4}(g)\) that can be attained before the gas will liquefy at \(152^{\circ} \mathrm{C}\) . The boiling point for Nic CO) is \(42^{\circ} \mathrm{C}\) and the enthalpy of vaporization is 29.0 \(\mathrm{kJ} / \mathrm{mol} .\) [Hint: The phase change reaction and the corresponding equilibrium expression are \(\mathrm{Ni}(\mathrm{CO})_{4}(l) \rightleftharpoons \mathrm{Ni}(\mathrm{CO})_{4}(g) \quad K=P_{\mathrm{NiCO} 4}\) greater than the \(K\) value. \(]\)

At \(100 .^{\circ} \mathrm{C}\) and \(1.00 \mathrm{atm}, \Delta H^{\circ}=40.6 \mathrm{kJ} / \mathrm{mol}\) for the vaporization of water. Estimate \(\Delta G^{\circ}\) for the vaporization of water at \(90 .^{\circ} \mathrm{C}\) and \(110 .^{\circ} \mathrm{C}\) . Assume \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) at \(100 .^{\circ} \mathrm{C}\) and 1.00 \(\mathrm{atm}\) do not depend on temperature.

Using Appendix 4 and the following data, determine \(S^{\circ}\) for \(\mathrm{Fe}(\mathrm{CO})_{5}(g) . \)\mathrm{Fe}(s)+5 \mathrm{CO}(g) \longrightarrow \mathrm{Fe}(\mathrm{CO})_{5}(g) \quad \Delta S^{\circ}=?\( \)\mathrm{Fe}(\mathrm{CO})_{5}(l) \longrightarrow \mathrm{Fe}(\mathrm{CO})_{5}(g) \quad \Delta S^{\circ}=107 \mathrm{J} / \mathrm{K}\( \)\mathrm{Fe}(s)+5 \mathrm{CO}(g) \longrightarrow \mathrm{Fe}(\mathrm{CO})_{5}(l) \quad \Delta S^{\circ}=-677 \mathrm{J} / \mathrm{K}$

For the reaction $$\mathrm{CS}_{2}(g)+3 \mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)+2 \mathrm{SO}_{2}(g)$$ \(\Delta S^{\circ}\) is equal to \(-143 \mathrm{JK}\) . Use this value and data from Appendix 4 to calculate the value of \(S^{\circ}\) for \(\mathrm{CS}_{2}(g) .\)

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