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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) \(\mathrm{P}_{2} \mathrm{O}_{5}\)

Short Answer

Expert verified
The conjugate base of \( \mathrm{H}_{2} \mathrm{PO}_{4}^{-} \) is \( \mathrm{HPO}_{4}^{2-} \). Option (a).

Step by step solution

01

Understand the Concept

The conjugate base of an acid is what is formed when the acid donates a proton (H鈦). For a given acid-base reaction, the acid will lose one hydrogen ion, resulting in a conjugate base.
02

Identify the Acid Species

In this exercise, the given acid species is \( \mathrm{H}_{2} \mathrm{PO}_{4}^{-} \). This is dihydrogen phosphate, which is polyatomic and has an overall one negative charge.
03

Apply the Proton Loss Rule

To find the conjugate base, we must remove one H鈦 ion from \( \mathrm{H}_{2} \mathrm{PO}_{4}^{-} \). This will decrease the number of hydrogens and increase the negative charge by one.
04

Write the Chemical Formula for the Conjugate Base

After removing one proton, \( \mathrm{H}_{2} \mathrm{PO}_{4}^{-} \) becomes \( \mathrm{HPO}_{4}^{2-} \). This molecule now has one less hydrogen and one more negative charge, indicating it is the conjugate base.
05

Verify with the Options

Match \( \mathrm{HPO}_{4}^{2-} \) with the provided choices. Choice (a), \( \mathrm{HPO}_{4}^{2-} \), matches our result, indicating it is the correct conjugate base.

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Key Concepts

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

Conjugate Base
When an acid donates a proton or hydrogen ion (H鈦), the remainder left is known as the conjugate base. This concept is crucial in understanding acid-base reactions. The acid loses one hydrogen ion, resulting in a conjugate base that carries one more negative charge than the acid it originated from.
A practical example is the dihydrogen phosphate ion ( H鈧侾O鈧勨伝). When this ion donates a proton, it transforms into the hydrogen phosphate ion ( HPO鈧劼测伝), which is the conjugate base. Knowing how to determine conjugate bases is an essential skill in chemistry, as it helps predict the behavior of acids and bases in reactions.
To find a conjugate base:
  • Identify the acid.
  • Remove a proton (H鈦) from the acid's formula.
  • Add one negative charge due to the loss of the proton.

By following these steps, you can accurately determine the conjugate base of any given acid.
Proton Transfer Reactions
In chemistry, proton transfer reactions are fundamental processes where a proton is transferred from one molecule to another. These reactions are at the heart of acid-base chemistry. Understanding proton transfer reactions helps in appreciating how acids and bases interact.
In an acid-base reaction:
  • The acid donates a proton.
  • The base accepts the proton.

For example, when dihydrogen phosphate ion ( H鈧侾O鈧勨伝) acts as an acid and donates a proton, it forms a conjugate base ( HPO鈧劼测伝). The molecule that receives the proton becomes the conjugate acid.
Proton transfer plays a crucial role in reactions in aqueous solutions. This type of reaction is reversible, and the direction may depend on the relative strengths of the acids and bases involved. To predict the direction and equilibrium position of these reactions, chemists often use the concept of pH and pKa values.
Dihydrogen Phosphate Ion
The dihydrogen phosphate ion ( H鈧侾O鈧勨伝) is a polyatomic ion with an overall charge of -1. It is an example of an ion that can participate in both donating and accepting protons, reflecting its amphiprotic nature.
Key features of the dihydrogen phosphate ion:
  • Contains two hydrogen atoms, one phosphorus atom, and four oxygen atoms.
  • Can act as an acid by donating a hydrogen ion to form HPO鈧劼测伝.
  • Capable of accepting a proton to become H鈧働O鈧.

The versatility of dihydrogen phosphate ions allows them to be central to the buffer systems in biological and chemical processes. Buffers help maintain a stable pH in solutions, which is important for numerous biological functions.
Understanding the behavior of dihydrogen phosphate ions in various reactions provides insight into their importance in chemistry, especially in biochemistry, where phosphate-based groups are prevalent.

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Most popular questions from this chapter

\(500 \mathrm{ml}\) of \(0.2 \mathrm{M} \mathrm{HCl}\) is mixed with \(500 \mathrm{ml}\) of \(0.2 \mathrm{M}\) \(\mathrm{CH}_{3} \mathrm{COOH} .25 \mathrm{ml}\) of the mixture is titrated with \(0.1\) M \(\mathrm{NaOH}\) solution. By how many units does the \(\mathrm{pH}\) change from the start to the stage when \(\mathrm{HCl}\) is just completely neutralized. \(\mathrm{K}_{\text {, for acetic acid }}=2.0 \times 10^{-5}\). (a) \(3.7\) (b) \(4.4\) (c) \(2.0\) (d) \(3.0\)

When \(60 \mathrm{ml}\) of \(0.1 \mathrm{M} \mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2}\) is mixed with \(40 \mathrm{ml}\) of \(0.125 \mathrm{M} \mathrm{Na}_{2} \mathrm{CO}_{3}, \mathrm{CaCO}_{3}\) precipitates. If \(\mathrm{K}_{\text {sp }}\) of \(\mathrm{CaCO}_{3}\) is \(5 \times 10^{-9} \mathrm{M}^{2}\), the \(\left[\mathrm{CO}_{3}^{2-}\right]\) in the resulting solution is (a) \(5 \times 10^{-8} \mathrm{M}\) (b) \(5 \times 10^{-9} \mathrm{M}\) (c) \(5 \times 10^{-6} \mathrm{M}\) (d) \(5 \times 10^{-7} \mathrm{M}\)

At what concentration of \(\mathrm{CH}_{3} \mathrm{COOH}\) will the \(\left[\mathrm{H}^{+}\right]\) obtained will be same as that obtained from \(10^{-2} \mathrm{M}\) \(\mathrm{HCOOH},\left(\mathrm{Ka}\left(\mathrm{CH}_{3} \mathrm{COOH}\right)=10^{-5}, \mathrm{Ka}(\mathrm{HCOOH})=10^{-4}\right)\) (a) \(10 \mathrm{M}\) (b) \(5 \mathrm{M}\) (c) \(10^{-1} \mathrm{M}\) (d) \(6 \mathrm{M}\)

Equilibrium constant of \(\mathrm{NH}_{4}^{+}\)to \(\mathrm{NH}_{3}\) and \(\mathrm{H}^{+}\)is \(10^{-10}\). The rate constant for \(\mathrm{NH}_{4}^{+}+\mathrm{OH}^{-} \rightarrow \mathrm{NH}_{3}+\mathrm{H}_{2} \mathrm{O}\) is \(10^{10} .\) The rate constant for \(\mathrm{NH}_{3}+\mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{NH}_{4}^{+}+\mathrm{OH}^{-}\)is (a) \(10^{5}\) (b) \(10^{6}\) (c) \(10^{8}\) (d) \(10^{9}\)

Which of the following solution(s) have \(\mathrm{pH}\) between 6 and \(7 ?\) 1\. \(2 \times 10^{-6} \mathrm{M} \mathrm{NaOH}\) 2\. \(2 \times 10^{-6} \mathrm{M} \mathrm{HCl}\) 3\. \(10^{-8} \mathrm{M} \mathrm{HCl}\) 4\. \(10^{-13} \mathrm{M} \mathrm{NaOH}\) (a) 1,2 (b) 2,3 (c) 3,4 (d) \(2,3,4\)

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