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If a substance is a Lewis acid, is it necessarily a Brønsted-Lowry acid? Is it necessarily an Arrhenius acid? Explain.

Short Answer

Expert verified
A substance being a Lewis acid, which is an electron-pair acceptor, does not necessarily mean it is a Brønsted-Lowry acid or an Arrhenius acid. The key differences are that Brønsted-Lowry acids donate protons in reactions, and Arrhenius acids increase the concentration of H⁺ ions in aqueous solutions. Some Lewis acids may not exhibit these behaviors, making it possible for a substance to be a Lewis acid without being a Brønsted-Lowry or Arrhenius acid.

Step by step solution

01

Defining Lewis, Brønsted-Lowry, and Arrhenius acids

A Lewis acid is a substance that can accept an electron pair from a Lewis base to form a coordinate covalent bond. In other words, it is an electron-pair acceptor. A Brønsted-Lowry acid is a substance that can donate a proton (H⁺) to a Brønsted-Lowry base in a reaction. So, it is a proton donor. An Arrhenius acid is a substance that, when dissolved in water, increases the concentration of H⁺ ions (or H₃O⁺, hydronium ions) in the solution. These are typically compounds containing hydrogen, such as hydrogen halides (e.g., HCl, HBr, HI).
02

Comparing Lewis acids to Brønsted-Lowry acids

Now that we have defined each type of acid, let's compare Lewis acids to Brønsted-Lowry acids. While both types of acids can accept electron pairs to form coordinate covalent bonds, the key difference is that Brønsted-Lowry acids include the donation of a proton in the reaction. It is possible for a substance to be a Lewis acid without being a Brønsted-Lowry acid, as it may not necessarily donate a proton during the reaction. Therefore, a substance being a Lewis acid does not necessarily mean it is a Brønsted-Lowry acid.
03

Comparing Lewis acids to Arrhenius acids

Next, let's compare Lewis acids to Arrhenius acids. Arrhenius acids specifically increase the concentration of H⁺ ions in an aqueous solution. Some Lewis acids may not necessarily do this, as they may accept electron pairs from Lewis bases without releasing H⁺ ions into the solution. Therefore, a substance being a Lewis acid does not necessarily mean it is an Arrhenius acid.
04

Conclusion

In conclusion, a substance being a Lewis acid is not sufficient to guarantee that it is also a Brønsted-Lowry or an Arrhenius acid. The key differences lie in whether the substance donates a proton (for Brønsted-Lowry) or increases the concentration of H⁺ ions in an aqueous solution (for Arrhenius).

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Most popular questions from this chapter

Predict the products of the following acid-base reactions, and predict whether the equilibrium lies to the left or to the right of the equation: (a) \(\mathrm{O}^{2-}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons\) (b) \(\mathrm{CH}_{3} \mathrm{COOH}(a q)+\mathrm{HS}^{-}(a q) \rightleftharpoons\) (c) \(\mathrm{NO}_{2}^{-}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons\)

Based on their compositions and structures and on conjugate acid-base relationships, select the stronger base in each of the following pairs: (a) \(\mathrm{NO}_{3}^{-}\) or \(\mathrm{NO}_{2}^{-}\), (b) \(\mathrm{PO}_{4}{ }^{3-}\) or \(\mathrm{AsO}_{4}{ }^{3-}\), (c) \(\mathrm{HCO}_{3}^{-}\) or \(\mathrm{CO}_{3}{ }^{2-}\)

For solutions of a weak acid, a graph of \(\mathrm{pH}\) versus the log of the initial acid concentration should be a straight line. What is the magnitude of the slope of that line?

(a) The hydrogen oxalate ion \(\left(\mathrm{HC}_{2} \mathrm{O}_{4}{ }^{-}\right)\) is amphiprotic. Write a balanced chemical equation showing how it acts as an acid toward water and another equation showing how it acts as a base toward water. (b) What is the conjugate acid of \(\mathrm{HC}_{2} \mathrm{O}_{4}^{-} ?\) What is its conjugate base?

Calculate \(\left[\mathrm{OH}^{-}\right]\) and \(\mathrm{pH}\) for (a) \(1.5 \times 10^{-3} \mathrm{M} \mathrm{Sr}(\mathrm{OH})_{2}\) (b) \(2.250 \mathrm{~g}\) of \(\mathrm{LiOH}\) in \(250.0 \mathrm{~mL}\) of solution, \((c) 100 \mathrm{~mL}\) of \(0.175 \mathrm{M} \mathrm{NaOH}\) diluted to \(2.00 \mathrm{~L},(\mathrm{~d})\) a solution formed by adding \(5.00 \mathrm{~mL}\) of \(0.105 \mathrm{M} \mathrm{KOH}\) to \(15.0 \mathrm{~mL}\) of \(9.5 \times 10^{-2} \mathrm{M} \mathrm{Ca}(\mathrm{OH})_{2}\)

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