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(a) Give the conjugate base of the following Bronsted-Lowry acids: (i) \(\mathrm{HIO}_{3},(\mathbf{i} \mathbf{i}) \mathrm{NH}_{4}^{+} .(\mathbf{b})\) Give the conjugate acid of the following Bronsted-Lowry bases: (i) \(\mathrm{O}^{2-},(\mathbf{i} \mathbf{i}) \mathrm{H}_{2} \mathrm{PO}_{4}^{-}\)

Short Answer

Expert verified
The conjugate bases and acids of the given Bronsted-Lowry acids and bases are: (a) Conjugate bases: (i) HIO鈧 鉃 IO鈧冣伝 (ii) NH鈧勨伜 鉃 NH鈧 (b) Conjugate acids: (i) O虏鈦 鉃 OH鈦 (ii) H鈧侾O鈧勨伝 鉃 H鈧働O鈧

Step by step solution

01

Identify the acid as HIO鈧

The given Bronsted-Lowry acid is HIO鈧.
02

Remove a proton from the acid

To find the conjugate base, we need to remove a proton (H鈦) from the given acid, which means we are left with IO鈧冣伝.
03

Write down the conjugate base

The conjugate base of HIO鈧 is IO鈧冣伝. (ii) NH鈧勨伜
04

Identify the acid as NH鈧勨伜

The given Bronsted-Lowry acid is NH鈧勨伜.
05

Remove a proton from the acid

To find the conjugate base, we need to remove a proton (H鈦) from the given acid, which means we are left with NH鈧.
06

Write down the conjugate base

The conjugate base of NH鈧勨伜 is NH鈧. (b) Give the conjugate acid of the following Bronsted-Lowry bases: (i) O虏鈦
07

Identify the base as O虏鈦

The given Bronsted-Lowry base is O虏鈦.
08

Add a proton to the base

To find the conjugate acid, we need to add a proton (H鈦) to the given base, which means we end up with OH鈦.
09

Write down the conjugate acid

The conjugate acid of O虏鈦 is OH鈦. (ii) H鈧侾O鈧勨伝
10

Identify the base as H鈧侾O鈧勨伝

The given Bronsted-Lowry base is H鈧侾O鈧勨伝.
11

Add a proton to the base

To find the conjugate acid, we need to add a proton (H鈦) to the given base, which means we end up with H鈧働O鈧.
12

Write down the conjugate acid

The conjugate acid of H鈧侾O鈧勨伝 is H鈧働O鈧.

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Key Concepts

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

Conjugate Base
In the world of Bronsted-Lowry acids and bases, understanding the concept of a conjugate base is central. When an acid donates a proton, or hydrogen ion ( H^+ ), it transforms into its conjugate base. This transformation is crucial to acid-base reactions. For example, when HIO_3 , an acid, loses a proton, it becomes the conjugate base, IO_3^- . Similarly, NH_4^+ , when losing a proton, turns into NH_3 , its conjugate base.

To recognize a conjugate base:
  • Start by identifying the acid in the reaction.
  • Remove one proton from this acid.
  • The resulting species is the conjugate base.
This process showcases the reversible nature of acid-base reactions, where the conjugate base can often absorb a proton to revert to the original acid, highlighting the complementary relationship between acids and bases.
Conjugate Acid
Just as acids have conjugate bases, bases have conjugate acids. The concept of a conjugate acid comes into play when a base accepts a proton, resulting in a newly formed acid. Consider the base O^{2-} ; when it gains a proton, it transforms into the conjugate acid OH^- . Another example is H_2PO_4^- , which, upon accepting a proton, becomes H_3PO_4 .

To understand conjugate acids, follow these steps:
  • Identify the base involved in the reaction.
  • Add one proton to the identified base.
  • The new species formed is the conjugate acid.
This exchange signifies the dynamic nature of acid-base chemistry, where bases turn into acids through proton acceptance, showcasing the balance and interplay between conjugate acid-base pairs.
Proton Transfer
At the heart of Bronsted-Lowry acid-base theory is the concept of proton transfer. This process involves the movement of a proton from an acid to a base. Such transfers are fundamental to many chemical reactions and drive the formation of conjugate acid-base pairs.

Consider the transformation of HIO_3 to IO_3^- and NH_4^+ to NH_3 , where protons are transferred from these acids, converting them to their respective conjugate bases. Similarly, the transition of O^{2-} to OH^- and H_2PO_4^- to H_3PO_4 involves an addition of protons, forming conjugate acids.

Important points to remember about proton transfer:
  • Proton transfer creates two new entities: a conjugate base and a conjugate acid.
  • It demonstrates the reversible reactions where an acid becomes a base and vice versa.
Mastering proton transfer helps in understanding the nuanced interactions in acid-base chemistry, forming the basis for predicting reaction outcomes and understanding their equilibrium states.

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Most popular questions from this chapter

codeine \(\left(\mathrm{C}_{18} \mathrm{H}_{21} \mathrm{NO}_{3}\right)\) is a weak organic base. \(\mathrm{A} 5.0 \times 10^{-3} \mathrm{M}\) solution of codeine has a pH of \(9.95 .\) Calculate the value of \(K_{b}\) for this substance. What is the \(\mathrm{pK}_{b}\) for this base?

Predict which member of each pair produces the more acidic aqueous solution: (a) \(\mathrm{K}^{+}\) or \(\mathrm{Cu}^{2+},(\mathbf{b}) \mathrm{Fe}^{2+}\) or \(\mathrm{Fe}^{3+}\) , (c) \(\mathrm{Al}^{3+}\) or \(\mathrm{Ga}^{3+}\) .

Pyridinium bromide \(\left(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{NHBr}\right)\) is a strong electrolyte that dissociates completely into \(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{NH}^{+}\) and \(\mathrm{Br}^{-} .\) An aqueous solution of pyridinium bromide has a pH of \(2.95 .\) (a) Write out the reaction that leads to this acidic pH. (b) Using Appendix D, calculate the \(K_{a}\) for pyridinium bromide. (c) A solution of pyridinium bromide has a pH of 2.95 . What is the concentration of the pyridinium cation atequilibrium, in units of molarity?

Butyric acid is responsible for the foul smell of rancid butter. The \(\mathrm{pK}_{a}\) of butyric acid is 4.84 (a) Calculate the pK \(_{b}\) for the butyrate ion. (b) Calculate the pH of a 0.050 \(M\) solution of butyric acid. (c) Calculate the pH of a 0.050\(M\) solution of sodium butyrate.

Indicate whether each of the following statements is true or false. For each statement that is false, correct the statement to make it true. (a) In general, the acidity of binary acids increases from left to right in a given row of the periodic table. (b) In a series of acids that have the same central atom, acid strength increases with the number of hydrogen atoms bonded to the central atom. (c) Hydrotelluric acid \(\left(\mathrm{H}_{2} \mathrm{Te}\right)\) is a stronger acid than \(\mathrm{H}_{2} \mathrm{S}\) because Te is more electronegative than \(\mathrm{S} .\)

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