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Identify the Br酶nsted-Lowry acid and the Br酶nsted-Lowry base on the left side of each of the following equations, and also identify the conjugate acid and conjugate base of each on the right side: (a) \(\mathrm{NH}_{4}^{+}(a q)+\mathrm{CN}^{-}(a q) \rightleftharpoons \mathrm{HCN}(a q)+\mathrm{NH}_{3}(a q)\) (b) \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{~N}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons\) $$ \left(\mathrm{CH}_{3}\right)_{3} \mathrm{NH}^{+}(a q)+\mathrm{OH}^{-}(a q) $$ (c) \(\mathrm{HCOOH}(a q)+\mathrm{PO}_{4}^{3-}(a q) \rightleftharpoons\) $$ \mathrm{HCOO}^{-}(a q)+\mathrm{HPO}_{4}^{2-}(a q) $$

Short Answer

Expert verified
(a) Acid: NH鈧勨伜, Base: CN鈦, Conjugate acid: HCN, Conjugate base: NH鈧. (b) Acid: H鈧侽, Base: (CH鈧)鈧僋, Conjugate acid: (CH鈧)鈧僋H鈦, Conjugate base: OH鈦. (c) Acid: HCOOH, Base: PO鈧劼斥伝, Conjugate acid: HPO鈧劼测伝, Conjugate base: HCOO鈦.

Step by step solution

01

Understanding Br酶nsted-Lowry Acid and Base

The Br酶nsted-Lowry theory defines an acid as a species that donates a proton (H鈦), and a base as a species that accepts a proton. In a chemical reaction, we identify which reactant gives away an H鈦 (acid) and which one gains an H鈦 (base).
02

Analyzing Reaction (a)

For \( \mathrm{NH}_{4}^{+}(a q) + \mathrm{CN}^{-}(a q) \rightleftharpoons \mathrm{HCN}(a q) + \mathrm{NH}_{3}(a q) \),NH鈧勨伜 donates a proton becoming NH鈧, thus it is the acid. CN鈦 accepts a proton to become HCN, thus it is the base.Conjugate pairs:- Conjugate acid: HCN- Conjugate base: NH鈧.
03

Analyzing Reaction (b)

For\( \left(\mathrm{CH}_{3}\right)_{3} \mathrm{~N}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \left(\mathrm{CH}_{3}\right)_{3} \mathrm{NH}^{+}(a q)+\mathrm{OH}^{-}(a q) \),water donates a proton to (CH鈧)鈧僋, making it the acid, and (CH鈧)鈧僋 accepts the proton, making it the base.Conjugate pairs:- Conjugate acid: (CH鈧)鈧僋H鈦- Conjugate base: OH鈦.
04

Analyzing Reaction (c)

For\( \mathrm{HCOOH}(a q)+\mathrm{PO}_{4}^{3-}(a q) \rightleftharpoons\mathrm{HCOO}^{-}(a q)+\mathrm{HPO}_{4}^{2-}(a q) \),HCOOH donates a proton to the phosphate ion, so it is the acid, and PO鈧劼斥伝 accepts the proton, so it is the base.Conjugate pairs:- Conjugate acid: HPO鈧劼测伝- Conjugate base: HCOO鈦.

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Key Concepts

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

Conjugate Acid
In the Br酶nsted-Lowry acid-base theory, the concept of a conjugate acid is central to understanding reactions involving proton transfers. A conjugate acid is formed when a base gains a proton ( H^+ ). This change occurs on the right side of the chemical equation, signifying the product side in an acid-base reaction.
When we look at reaction (a) from the problem statement, CN^- gains a proton from NH鈧刕+, transforming into HCN. Thus, HCN is the conjugate acid of CN^-. Similarly, in reaction (b), the base (CH鈧)鈧僋 accepts a proton from water ( H鈧侽 ), resulting in the formation of (CH鈧)鈧僋H^+, the conjugate acid in this scenario.
For reaction (c), the base PO鈧刕{3-} accepts a proton from HCOOH, forming HPO鈧刕{2-}, which acts as the conjugate acid. Always remember, identifying the conjugate acid involves recognizing the proton accepted by the base, creating a new acid-product pair.
Conjugate Base
Conversely, the conjugate base is formed when an acid donates a proton in an acid-base reaction. This process results in the acid turning into its conjugate base, observable on the reaction's right side.
In reaction (a), NH鈧刕+ acts as an acid and donates a proton, transforming into NH鈧. Thus, NH鈧 is the conjugate base that results from this proton donation. Looking at reaction (b), water ( H鈧侽 ) acts as the acid. After losing a proton to (CH鈧)鈧僋, it turns into OH^-, its conjugate base.
In reaction (c), HCOOH is the acid that donates a proton to PO鈧刕{3-}, resulting in the formation of HCOO^-. Consequently, HCOO^- is the conjugate base here. This conversion highlights how acids transform into their conjugate bases by losing protons.
Proton Transfer
Proton transfer is the essence of any Br酶nsted-Lowry acid-base reaction. It involves the movement of a proton ( H^+ ) from an acid to a base. This movement triggers the formation of the conjugate acid and conjugate base, respectively.
During this process, the material that loses the proton is called the Br酶nsted-Lowry acid, while the one that gains it is called the Br酶nsted-Lowry base. In our example reactions, like in (a) NH鈧刕+ donates a proton to CN^-, and in (b) water donates to (CH鈧)鈧僋. Both instances illustrate proton transfer.
Understanding which component donates and which one accepts the proton is key to correctly identifying acids and bases in reactions, making proton transfer a crucial step in comprehending acid-base dynamics.
Acid-Base Reaction
An acid-base reaction in the Br酶nsted-Lowry framework is characterized by the exchange of protons. This type of chemical reaction centers around one species acting as an acid by donating a proton, and another acting as a base by accepting it.
In examining reactions like (a) NH鈧刕+ + CN^- ightleftharpoons HCN + NH鈧 , you'll see NH鈧刕+ donating a proton to CN^-, creating NH鈧 and HCN as by-products. Each acid-base reaction results in the formation of a conjugate acid and a conjugate base.
Mastering acid-base reactions means being able to identify who donates and who accepts protons. This understanding helps in predicting the behavior of acids and bases in different chemical environments. It's all about recognizing the shift in proton ownership!

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Most popular questions from this chapter

The hypochlorite ion, \(\mathrm{ClO}^{-}\), acts as a weak base. (a) Is \(\mathrm{ClO}^{-}\) a stronger or weaker base than hydroxylamine? (b) When \(\mathrm{ClO}^{-}\) acts as a base, which atom, \(\mathrm{Cl}\) or \(\mathrm{O},\) acts as the proton acceptor? (c) Can you use formal charges to rationalize your answer to part (b)?

Calculate the pH of a solution made by adding \(1.00 \mathrm{~g}\) potassium oxide \(\left(\mathrm{K}_{2} \mathrm{O}\right)\) to enough water to make \(2.00 \mathrm{~L}\) of solution.

Write the chemical equation and the \(K_{a}\) expression for the acid dissociation of each of the following acids in aqueous solution. First show the reaction with \(\mathrm{H}^{+}(a q)\) as a product and then with the hydronium ion: \((\mathbf{a}) \mathrm{HSO}_{4}^{-}\), (b) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{OH}\).

(a) Given that \(K_{b}\) for ammonia is \(1.8 \times 10^{-5}\) and that for hydroxylamine is \(1.1 \times 10^{-8}\), which is the stronger base? (b) Which is the stronger acid, the ammonium ion or the hydroxylammonium ion? (c) Calculate \(K_{a}\) values for \(\mathrm{NH}_{4}^{+}\) and \(\mathrm{H}_{3} \mathrm{NOH}^{+}\).

Calculate the \(\mathrm{pH}\) of each of the following strong acid solutions: (a) \(0.0178 \mathrm{M} \mathrm{HNO}_{3},\) (b) \(0.500 \mathrm{~g}\) of \(\mathrm{HClO}_{3}\) in \(5.00 \mathrm{~L}\) of solution, \((\mathbf{c}) 5.00 \mathrm{~mL}\) of \(2.00 \mathrm{M} \mathrm{HCl}\) diluted to \(0.500 \mathrm{~L}\), (d) a mixture formed by adding \(75.0 \mathrm{~mL}\) of \(0.010 \mathrm{M} \mathrm{HCl}\) to \(125 \mathrm{~mL}\) of \(0.020 \mathrm{M}\) HBr.

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