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An acid \((\mathrm{HA})\) can be defined as a substance that donates a proton to a base, for example water. The protondonation reaction usually is an equilibrium reaction and is written as $$ \mathrm{H}: \mathrm{A}+\mathrm{H}: \ddot{\mathrm{O}}: \mathrm{H} \rightleftarrows \mathrm{H}: \stackrel{\mathrm{H}}{\mathrm{O}} \cdot \stackrel{\oplus}{\mathrm{H}}+: \AA^{\ominus} $$ Predict which member of each of the following pairs of compounds would be the stronger acid. Give your reasons. a. \(\mathrm{LiH}, \mathrm{HF}\) b. \(\mathrm{NH}_{3}, \mathrm{H}_{2} \mathrm{O}\) c. \(\mathrm{H}_{2} \mathrm{O}_{2}, \mathrm{H}_{2} \mathrm{O}\) d. \(\mathrm{CH}_{4}, \mathrm{CF}_{3} \mathrm{H}\)

Short Answer

Expert verified
HF, H2O, H2O2, and CF3H are the stronger acids.

Step by step solution

01

Assess with HF and LiH

To determine which is a stronger acid between HF and LiH, consider the ability to donate a proton. HF is a polar covalent bond with a highly electronegative fluorine, allowing it to easily release the proton (H+). LiH is an ionic compound with hydrogen as a hydride ion (H-), so it is not an acid; instead, it tends to gain a proton. Therefore, HF is a stronger acid than LiH.
02

Compare NH3 and H2O

Analyzing the basic and acidic properties, water (H2O) can act as an acid when donating a proton to form OH-. Ammonia (NH3), on the other hand, is more basic due to the nitrogen atom, which is inclined to accept protons rather than donate them. Hence, H2O is a stronger acid compared to NH3.
03

Evaluate H2O2 vs H2O

Hydrogen peroxide (H2O2) is a stronger acid compared to water (H2O) due to the presence of an additional oxygen, which increases the molecule's oxidizing potential and stabilizes the resultant negative charge after donating a proton. This makes H2O2 a slightly stronger acid than H2O.
04

Compare CH4 and CF3H

Methane (CH4) hardly acts as an acid because its carbon-hydrogen bond is quite non-polar and stable. In contrast, CF3H has an electron-withdrawing trifluoromethyl group (CF3) that increases the polarity of the C-H bond, facilitating easier proton donation. Therefore, CF3H is a stronger acid than CH4.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Proton Donation
Proton donation is a fundamental process in acid-base chemistry, where an acid, typically a compound containing hydrogen, donates a proton (hydrogen ion, \( \text{H}^+ \)) to a base. Imagine an acid as a helpful donor handing over a proton to a willing base. This exchange is what defines the acid's role in a chemical reaction. Let's look at water as a base, for example. When an acid like \( \text{HA} \) meets water \(( \text{H}_2 \text{O} )\), the water molecule accepts the hydrogen ion, resulting in \( \text{H}_3 \text{O}^+ \) (the hydronium ion), and leaving behind the conjugate base \( \text{A}^- \).

This donation process is important because it initiates the acid-base reaction and determines the acidity of the substance. Polarity and electronegativity play crucial roles in how easily an acid can donate its proton. For example, in hydrofluoric acid (HF), the electronegative fluorine pulls electron density away from hydrogen, making it easier to donate that proton.
Acid Strength Comparison
Understanding the strength of acids involves examining their ability to donate protons efficiently. Always remember, a stronger acid donates its protons more readily than a weaker one.

In pairing compounds like HF and LiH, HF stands out as a stronger acid due to its polar covalent bond, which allows the hydrogen to be donated as a proton. On the contrary, LiH holds hydrogen as a hydride \(( \text{H}^- )\), so it does not behave as an acid.

Let's compare ammonia (\( \text{NH}_3 \)) and water (\( \text{H}_2\text{O} \)); water is the stronger acid. Despite both being capable of donating protons, water more readily forms hydronium \(( \text{H}_3 \text{O}^+ )\), given its molecular structure.

When comparing \( \text{H}_2\text{O}_2 \) and \( \text{H}_2\text{O} \), the additional oxygen in \( \text{H}_2\text{O}_2 \) increases its oxidizing ability and stabilizes the acid further after proton loss, thereby making it the stronger acid.

Lastly, \( \text{CH}_4 \) versus \( \text{CF}_3\text{H} \). Methane \(( \text{CH}_4 )\) has a very stable C-H bond, making it resist proton donation. Contrastingly, the presence of the trifluoromethyl group in \( \text{CF}_3\text{H} \) increases the bond's polarity, enhancing its ability to donate the hydrogen ion.
Equilibrium Reaction
Equilibrium reactions are a delicate balance in acid-base chemistry where chemical reactions reach a state where the rate of the forward reaction equals the rate of the reverse reaction. In the context of proton donation, an acid \(( \text{HA} )\) may donate a proton to a base like water, resulting in an equilibrium:

\[ \text{HA} + \text{H}_2\text{O} \rightleftharpoons \text{H}_3\text{O}^+ + \text{A}^- \]

At equilibrium, the concentrations of reactants and products remain constant, not because reactions have stopped, but because they are occurring at equal rates in both directions. This means that, at equilibrium, the tendency of the acid to donate a proton is perfectly balanced by the likelihood of the conjugate base reclaiming it.

Equilibrium constants \(( K_a )\) are used to express the strength of an acid; a larger \( K_a \) value signifies a stronger acid, indicating that the equilibrium is more shifted toward products, meaning more protons are donated. Recognizing these equilibriums helps in predicting how an acid behaves in a solution, indispensable for chemistry students to understand the stability and reactivity of acids.

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Most popular questions from this chapter

A gaseous compound of formula \(\mathrm{C}_{2} \mathrm{H}_{4}\) reacts with liquid bromine \(\left(\mathrm{Br}_{2}\right)\) to give a single \(\mathrm{C}_{2} \mathrm{H}_{4} \mathrm{Br}_{2}\) compound. The \(\mathrm{C}_{2} \mathrm{H}_{4} \mathrm{Br}_{2}\) so formed gives only one \(\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{Br}_{3}\) substitution product. Deduce the structure of \(\mathrm{C}_{2} \mathrm{H}_{4}\) and the bromo compounds derived from it. (This was a key problem for the early organic chemists.)

Lithium hydride could be written as either \(\mathrm{Li}^{\oplus}: \mathrm{H}^{\ominus}\) or \(\mathrm{H}^{\oplus}: \mathrm{Li}^{\ominus}\) depending on whether lithium or hydrogen is more electron-attracting. Explain why hydrogen is actually more electron-attracting, making the correct structure \(\mathrm{Li}^{\oplus}: \mathrm{H}^{\ominus}\).

The compound \(\mathrm{C}_{2} \mathrm{H}_{5}\) Br reacts slowly with the compound \(\mathrm{CH}_{4} \mathrm{O}\) to yield a single substance of formula \(\mathrm{C}_{3} \mathrm{H}_{8} \mathrm{O}\). Assuming normal valences throughout, write structural formulas for \(\mathrm{CH}_{4} \mathrm{O}\) and the three different possible structural (not rotational) isomers of \(\mathrm{C}_{3} \mathrm{H}_{8} \mathrm{O}\) and show how the principle of least structural change favors one of them as the reaction product. What would you expect to be formed from each of these three \(\mathrm{C}_{3} \mathrm{H}_{8} \mathrm{O}\) isomers with strong hydrobromic acid?

A compound of formula \(\mathrm{C}_{5} \mathrm{H}_{12}\) gives only a single monobromo substitution product of formula \(\mathrm{C}_{5} \mathrm{H}_{11} \mathrm{Br}\). What is the structure of this \(\mathrm{C}_{5} \mathrm{H}_{12}\) isomer? (Notice that carbon can form both continuous chains and branched chains. Also notice that structures such as the following represent the same isomer because the bonds to carbon are tetrahedral and are free to rotate.)

Dimethylmercury, \(\mathrm{CH}_{3}-\mathrm{Hg}-\mathrm{CH}_{3}\), is a volatile compound of bp \(96^{\circ}\), whereas mercuric fluoride \(\mathrm{F}-\mathrm{Hg}-\mathrm{F}\) is a high-melting solid having mp \(570^{\circ} .\) Explain what differences in bonding in the two substances are expected that can account for the great differences in physical properties.

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