Problem 266
Question
The conjugate base of \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\)is (a) \(\mathrm{HPO}_{4}^{2-}\) (b) \(\mathrm{H}_{3} \mathrm{PO}_{4}\) (c) \(\mathrm{PO}_{4}^{3-}\) (d) \(\mathrm{P}_{2} \mathrm{O}_{5}\)
Step-by-Step Solution
Verified Answer
The conjugate base of \\(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\\) is \\(\mathrm{HPO}_{4}^{2-}\\) (Option a).
1Step 1: Understanding the Concept
In chemistry, a conjugate base is formed when an acid donates a proton (H⁺). The species that remains after the acid loses its proton is the conjugate base of that acid.
2Step 2: Identify the Original Acid
The acid in question is \( \mathrm{H}_{2} \mathrm{PO}_{4}^{-} \) which is known as dihydrogen phosphate.
3Step 3: Remove a Proton
Since forming a conjugate base involves removing a proton (H⁺) from the acid, remove one hydrogen ion from \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\). This leaves us with \(\mathrm{HPO}_{4}^{2-}\).
4Step 4: Determine the Charge
When \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\) loses a proton, the charge of the base must be updated. Originally, it is -1, and after losing one H⁺, it becomes \(\mathrm{HPO}_{4}^{2-}\).
5Step 5: Confirm the Conjugate Base
Now confirm the conjugate base which is \(\mathrm{HPO}_{4}^{2-}\) among the options given: (a) \(\mathrm{HPO}_{4}^{2-}\) (b) \(\mathrm{H}_{3} \mathrm{PO}_{4}\) (c) \(\mathrm{PO}_{4}^{3-}\) (d) \(\mathrm{P}_{2} \mathrm{O}_{5}\)Since \(\mathrm{HPO}_{4}^{2-}\) is derived from \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\) by losing an H⁺, the correct answer is (a).
Key Concepts
Acid-Base ReactionsDihydrogen PhosphateProton Transfer
Acid-Base Reactions
Acid-base reactions are a fundamental concept in chemistry and involve the transfer of protons between molecules. In these reactions, an acid donates a proton (H⁺) to a base. The original acid transforms into its conjugate base after losing the proton, while the original base becomes a conjugate acid after gaining a proton.
It's essential to remember a few key points about acid-base reactions:
It's essential to remember a few key points about acid-base reactions:
- The strength of an acid is determined by its ability to donate a proton easily.
- The strength of a base is reflected in its ability to accept a proton.
- In these reactions, acids and bases always come in pairs known as conjugate acid-base pairs.
- When an acid donates a proton, the residual part is its conjugate base.
- When a base accepts a proton, the resulting species is its conjugate acid.
Dihydrogen Phosphate
Dihydrogen phosphate, written chemically as \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\), is a prevalent species in buffer solutions and plays a vital role in biological systems.
Here's what you should know about dihydrogen phosphate:
Here's what you should know about dihydrogen phosphate:
- It is an anion with a -1 charge due to its extra proton.
- Functioning as an acid, it can donate a proton to become \(\mathrm{HPO}_{4}^{2-}\), which is its conjugate base.
- It's part of the larger phosphate system involved in energy transfer processes like ATP in living organisms.
Proton Transfer
Proton transfer is a crucial component of many chemical reactions, particularly in the context of acid-base chemistry. It involves the movement of a hydrogen ion (H⁺) from one species to another.
In more detail, the process works as follows:
In more detail, the process works as follows:
- During a proton transfer, the acid donates a proton to the base.
- Proton transfer is typically rapid and often reversible, meaning the reaction can proceed both forward and backward.
- It results in changes in the charge and structure of the molecules involved.
- This mechanism is also a key part of biological systems, including enzyme activity and cellular energy management.
Other exercises in this chapter
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The molar solubility (in \(\mathrm{mol} \mathrm{L}^{-1}\) ) of a sparingly soluble salt \(\mathrm{MX}_{4}\) is 's'. The corresponding solubility product is Ksp.
View solution Problem 269
The conjugate base of \(\mathrm{OH}^{-}\)is [2005] (a) \(\mathrm{H}_{2} \mathrm{O}\) (b) \(\mathrm{O}^{2-}\) (c) \(\mathrm{O}^{-}\) (d) \(\mathrm{O}_{2}\)
View solution