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Chapter 14: Problem 31

Of the hydrogen halides, only HF is a weak acid. Give a possible explanation.

Short Answer

Expert verified
The strength of the H-X bond and electronegativity of halogens are key factors in determining hydrogen halide acidity. The HF bond is the strongest and fluorine has the highest electronegativity, resulting in difficult proton donation and making HF a weak acid. In contrast, other hydrogen halides have weaker H-X bonds and less electronegative halogens, causing them to nearly completely dissociate in water and be considered strong acids.

Step by step solution

01

1. Defining Acids and Acid Strength

An acid is a substance that donates a proton (H+) to a base, according to the Bronsted-Lowry definition. Acid strength is determined by the tendency of an acid to donate a proton, and a strong acid completely dissociates in water, while a weak acid only partially dissociates.
02

2. Factors Affecting Acid Strength

There are several factors that can affect the acidity of a compound, such as the electronegativity of the atoms involved, atom size, bond strength, and the stability of the resulting ion after the loss of a proton. In the case of hydrogen halides, we will focus on the bond strength and the electronegativity of the halogens.
03

3. Understanding Hydrogen Halides

Hydrogen halides are binary compounds in which a hydrogen atom (H) is bonded to a halogen atom (X), giving the general formula HX. The hydrogen halides are hydrogen fluoride (HF), hydrogen chloride (HCl), hydrogen bromide (HBr), and hydrogen iodide (HI).
04

4. The Electronegativity of Halogens

The electronegativity of an atom is its tendency to attract electrons. In general, electronegativity increases as you move across a period from left to right and decreases as you move down a group in the periodic table. Thus, the electronegativity of the halogens decreases as you descend the group: F > Cl > Br > I.
05

5. Bond Strength of Hydrogen Halides

The bond strength between the hydrogen and halogen atoms in hydrogen halides usually decreases as the size of the halogen atom increases. Small, highly electronegative atoms form strong bonds due to their ability to attract electrons, while large, less electronegative atoms form weaker bonds. The bond strength between hydrogen and the halogens is therefore in the following order: HF > HCl > HBr > HI.
06

6. Possible Explanation for HF Being a Weak Acid

The strength of the bond between hydrogen and halogens is a key factor in determining the acidity of hydrogen halides. Since the HF bond is the strongest, it is more difficult for HF to donate its proton than other hydrogen halides. Additionally, the high electronegativity of fluorine creates strong electrostatic attractions between the hydrogen and fluorine atoms, leading to a lesser tendency for the proton to be donated to another molecule or ion. As a result, HF only partially dissociates in water, making it a weak acid, while the other hydrogen halides are strong acids because they almost completely dissociate in water due to weaker H-X bonds and less electronegative halogens (X).

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Most popular questions from this chapter

A solution is prepared by adding \(50.0 \mathrm{~mL}\) concentrated hydrochloric acid and \(20.0 \mathrm{~mL}\) concentrated nitric acid to \(300 \mathrm{~mL}\) water. More water is added until the final volume is \(1.00 \mathrm{~L}\). Calculate \(\left[\mathrm{H}^{+}\right],\left[\mathrm{OH}^{-}\right]\), and the \(\mathrm{pH}\) for this solution. [Hint: Concentrated HCl is \(38 \%\) HCl (by mass) and has a density of \(1.19\) \(\mathrm{g} / \mathrm{mL} ;\) concentrated \(\mathrm{HNO}_{3}\) is \(70 . \% \mathrm{HNO}_{3}\) (by mass) and has a density of \(1.42 \mathrm{~g} / \mathrm{mL} .\) ]

Place the species in each of the following groups in order of increasing acid strength. a. \(\mathrm{H}_{2} \mathrm{O}, \mathrm{H}_{2} \mathrm{~S}, \mathrm{H}_{2} \mathrm{Se}\) (bond energies: \(\mathrm{H}-\mathrm{O}, 467 \mathrm{~kJ} / \mathrm{mol} ; \mathrm{H}-\mathrm{S}\), \(363 \mathrm{~kJ} / \mathrm{mol} ; \mathrm{H}-\mathrm{Se}, 276 \mathrm{~kJ} / \mathrm{mol})\) b. \(\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}, \mathrm{FCH}_{2} \mathrm{CO}_{2} \mathrm{H}, \mathrm{F}_{2} \mathrm{CHCO}_{2} \mathrm{H}, \mathrm{F}_{3} \mathrm{CCO}_{2} \mathrm{H}\) c. \(\mathrm{NH}_{4}^{+}, \mathrm{HONH}_{3}{ }^{+}\) d. \(\mathrm{NH}_{4}{ }^{+}, \mathrm{PH}_{4}{ }^{+}\) (bond energies: \(\mathrm{N}-\mathrm{H}, 391 \mathrm{~kJ} / \mathrm{mol} ; \mathrm{P}-\mathrm{H}, 322\) \(\mathrm{kJ} / \mathrm{mol}\) ) Give reasons for the orders you chose.

Calculate the \(\mathrm{pH}\) of a \(0.10 \mathrm{M} \mathrm{CoCl}_{3}\) solution. The \(K_{\mathrm{a}}\) value for \(\mathrm{Co}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}^{3+}\) is \(1.0 \times 10^{-5}\).

Calculate the mass of \(\mathrm{HONH}_{2}\) required to dissolve in enough water to make \(250.0 \mathrm{~mL}\) of solution having a \(\mathrm{pH}\) of \(10.00\left(K_{\mathrm{b}}=\right.\) \(\left.1.1 \times 10^{-8}\right)\)

Calculate the concentration of all species present and the \(\mathrm{pH}\) of a \(0.020 M \mathrm{HF}\) solution.
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