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Chapter 14: Problem 33

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

Short Answer

Expert verified
HF is a weak acid due to its strong bond strength, high electronegativity, and small atomic size. These factors make it more difficult for HF to dissociate in water and donate a proton (H+), leading to partial dissociation and its weak acid characteristic. On the other hand, other hydrogen halides (HCl, HBr, and HI) more readily dissociate in water, making them strong acids.

Step by step solution

01

Understand acid strength

A strong acid is an acid that completely dissociates in water, releasing protons (H+) and forming the conjugate base (A-). On the other hand, a weak acid is an acid that only partially dissociates in water, creating an equilibrium between the dissociated and undissociated forms. The strength of an acid can be determined by the ease of donating a proton (H+) and the stability of the conjugate base (A-). In the context of this question, we will focus on hydrogen halides and their strength of bonding as acids.
02

Consider the bond strength of hydrogen halides

The hydrogen halides are composed of a hydrogen atom (H) bonded to a halide atom (X) which belongs to group 7A in the periodic table (F, Cl, Br, I). The bond strength (or bond dissociation energy) of these H-X compounds typically decreases as we move down the group in the periodic table. This means that H-F has the strongest bond, followed by H-Cl, H-Br, and H-I. A strong bond between H and X could make it more difficult for H-X to lose a proton (H+).
03

Consider electronegativity and size of halide atoms

As we move down the halogens, the electronegativity decreases, and the atomic size increases. Since electronegativity is the measure of the ability of an element to attract electrons, the higher the electronegativity, the stronger the electron attraction. This means that HF has the highest electronegativity and a small atomic size, and this characteristic increases its bond strength.
04

Explain why HF is a weak acid

Considering the bond strength, electronegativity, and atomic size, we can now explain why HF is a weak acid compared to other hydrogen halides. The higher bond strength in HF (due to its higher electronegativity and smaller atomic size) makes it more difficult for the H-F bond to break. As a result, HF is less likely to donate a proton (H+) and dissociate into F- when placed in water. The partial dissociation of HF in water results in the weak acid characteristic, while the other hydrogen halides (HCl, HBr, and HI) more readily dissociate and are considered strong acids.

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Most popular questions from this chapter

Papaverine hydrochloride (abbreviated papH \(^{+} \mathrm{Cl}^{-} ;\) molar mass \(=378.85 \mathrm{~g} / \mathrm{mol}\) ) is a drug that belongs to a group of medicines called vasodilators, which cause blood vessels to expand, thereby increasing blood flow. This drug is the conjugate acid of the weak base papaverine (abbreviated pap; \(K_{\mathrm{b}}=\) \(8.33 \times 10^{-9}\) at \(35.0^{\circ} \mathrm{C}\) ). Calculate the \(\mathrm{pH}\) of a \(30.0-\mathrm{mg} / \mathrm{mL}\) aqueous dose of papH \(^{+} \mathrm{Cl}^{-}\) prepared at \(35.0^{\circ} \mathrm{C} . K_{\mathrm{w}}\) at \(35.0^{\circ} \mathrm{C}\) is \(2.1 \times 10^{-14}\).

The \(\mathrm{pH}\) of \(1.0 \times 10^{-8} \mathrm{M}\) hydrochloric acid is not \(8.00\). The correct \(\mathrm{pH}\) can be calculated by considering the relationship between the molarities of the three principal ions in the solution \(\left(\mathrm{H}^{+}, \mathrm{Cl}^{-}\right.\), and \(\left.\mathrm{OH}^{-}\right) .\) These molarities can be calculated from algebraic equations that can be derived from the considerations given below. a. The solution is electrically neutral. b. The hydrochloric acid can be assumed to be \(100 \%\) ionized. c. The product of the molarities of the hydronium ions and the hydroxide ions must equal \(K_{\mathrm{w}}\). Calculate the \(\mathrm{pH}\) of a \(1.0 \times 10^{-8}-\mathrm{M} \mathrm{HCl}\) solution.

Rank the following \(0.10 M\) solutions in order of increasing pH. a. \(\mathrm{NH}_{3}\) d. \(\mathrm{KCl}\) b. \(\mathrm{KOH}\) e. \(\mathrm{HCl}\) c. \(\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\)

Calculate the \(\mathrm{pH}\) of a \(5.0 \times 10^{-3}-M\) solution of \(\mathrm{H}_{2} \mathrm{SO}_{4}\).

Write out the stepwise \(K_{\mathrm{a}}\) reactions for citric acid \(\left(\mathrm{H}_{3} \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{O}_{7}\right)\), a triprotic acid.
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