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Chapter 17: Problem 20

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
HF is a weak acid because of its strong H-F bond, which requires more energy to break and results in a smaller increase in entropy upon dissociation in water compared to other hydrogen halides, such as HCl, HBr, and HI. This less favorable entropy change makes the ionization of HF less spontaneous, leading to its classification as a weak acid.

Step by step solution

01

Understand the concept of strong acids and weak acids

Strong acids completely ionize in water, donating all their H+ ions to the solution, while weak acids only partially ionize in water, meaning they donate only a fraction of their H+ ions. Hydrogen halides such as HCl, HBr, and HI are strong acids, while HF is a weak acid.
02

Define entropy

Entropy is a measure of disorder or randomness in a system. In general, entropy increases during reactions that create more disorder, for example, when a substance dissolves in water. Therefore, the extent to which an acid ionizes in water is dependent on the change in entropy in the system.
03

Examine the bond strength and size of atoms

In general, bond strength decreases as the size of the atoms involved in the bond increases. HF has the strongest bond among all hydrogen halides due to its small size (the smaller the atoms, the stronger the bond). On the other hand, HCl, HBr, and HI have weaker bonds as the size of Cl, Br, and I atoms increases down the halide family.
04

Compare the dissociation of HF and other hydrogen halides

When hydrogen halides dissociate in water, the H-X bond must break and H+ and X- ions form. In the case of HCl, HBr, and HI, this is a relatively easy process because of their weaker bonds, leading to a higher degree of ionization. However, for HF, the strong bond between the small H and F atoms makes this dissociation process less favorable (enthalpy-wise), resulting in a smaller amount of ionization and consequently, being a weak acid.
05

Relate the dissociation of HF and entropy

Due to the strong H-F bond, relatively more energy is required to break the bond, which means that the dissolution of HF leads to a smaller increase in entropy compared to other hydrogen halides. Since an increase in entropy generally favors spontaneity, the lesser increase in entropy due to HF dissociation makes it less spontaneous, hence causing it to be a weak acid.
06

Conclusion

HF is a weak acid and not a strong acid like other hydrogen halides mainly due to its strong H-F bond, resulting in less favorable entropy changes when HF dissociates in water. This causes HF to ionize less in water and therefore be considered a weak acid.

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Most popular questions from this chapter

Many biochemical reactions that occur in cells require relatively high concentrations of potassium ion \(\left(\mathrm{K}^{+}\right)\). The concentration of \(\mathrm{K}^{+}\) in muscle cells is about \(0.15 \mathrm{M}\). The concentration of \(\mathrm{K}^{+}\) in blood plasma is about \(0.0050 \mathrm{M}\). The high internal concentration in cells is maintained by pumping \(\mathrm{K}^{+}\) from the plasma. How much work must be done to transport \(1.0\) mole of \(\mathrm{K}^{+}\) from the blood to the inside of a muscle cell at \(37^{\circ} \mathrm{C}\), normal body temperature? When \(1.0\) mole of \(\mathrm{K}^{+}\) is transferred from blood to the cells, do any other ions have to be transported? Why or why not?

The equilibrium constant for a certain reaction increases by a factor of \(6.67\) when the temperature is increased from \(300.0 \mathrm{~K}\) to \(350.0 \mathrm{~K} .\) Calculate the standard change in enthalpy \(\left(\Delta H^{\circ}\right)\) for this reaction (assuming \(\Delta H^{\circ}\) is temperature-independent).

Given the following data: $$ \begin{aligned} 2 \mathrm{C}_{6} \mathrm{H}_{6}(l)+15 \mathrm{O}_{2}(g) \longrightarrow & 12 \mathrm{CO}_{2}(g)+& 6 \mathrm{H}_{2} \mathrm{O}(l) \\ & \Delta G^{\circ}=-6399 \mathrm{~kJ} \\ \mathrm{C}(s)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g) & \Delta G^{\circ}=-394 \mathrm{~kJ} \\ \mathrm{H}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l) & \Delta G^{\circ}=-237 \mathrm{~kJ} \end{aligned} $$ calculate \(\Delta G^{\circ}\) for the reaction $$ 6 \mathrm{C}(s)+3 \mathrm{H}_{2}(g) \longrightarrow \mathrm{C}_{6} \mathrm{H}_{6}(l) $$

Choose the substance with the larger positional probability in each case. a. 1 mole of \(\mathrm{H}_{2}\) (at STP) or 1 mole of \(\mathrm{H}_{2}\) (at \(\left.100^{\circ} \mathrm{C}, 0.5 \mathrm{~atm}\right)\) b. 1 mole of \(\mathrm{N}_{2}\) (at STP) or 1 mole of \(\mathrm{N}_{2}\) (at \(\left.100 \mathrm{~K}, 2.0 \mathrm{~atm}\right)\) c. 1 mole of \(\mathrm{H}_{2} \mathrm{O}(s)\) (at \(\left.0^{\circ} \mathrm{C}\right)\) or 1 mole of \(\mathrm{H}_{2} \mathrm{O}(l)\) (at \(\left.20^{\circ} \mathrm{C}\right)\)

Hydrogen cyanide is produced industrially by the following exothermic reaction: Is the high temperature needed for thermodynamic or kinetic reasons?
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