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Write the formula of the conjugate base in the reaction of each acid with water. (a) \(\mathrm{HIO}_{3} ;\) (b) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOH}\) (c) \(\mathrm{HPO}_{4}^{2-} ;\) (d) \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{3}^{+}\)

Short Answer

Expert verified
The formulas for the conjugate bases are (a) \( IO_{3}^{-} \); (b) \( C_{6}H_{5}COO^{-} \); (c) \( PO_{4}^{3-} \); (d) \( C_{2}H_{5}NH_{2} \)

Step by step solution

01

Write the Acid Reaction with Water

For each given acid, write an equation of it reacting with water. For example, the reaction of \(HIO_{3}\) with water can be written as: \(HIO_{3} + H_{2}O ↔ H_{3}O^{+} + IO_{3}^{-}\)
02

Identify the Conjugate Base

The conjugate base of the acid is the substance that remains after the acid donates a proton (H+). This will be obvious from the right side of the equation. For our example in step 1, \(IO_{3}^{-}\) is the conjugate base of the acid \(HIO_{3}\)
03

Repeat for the remaining acids

Repeat the above steps for each given acid: \(C_{6}H_{5}COOH\), \(HPO_{4}^{2-}\), and \(C_{2}H_{5}NH_{3}^{+}\). The conjugate bases obtained will be \(C_{6}H_{5}COO^{-}\), \(PO_{4}^{3-}\), and \(C_{2}H_{5}NH_{2}\) respectively.
04

Validate the Results

For each equation outlined in the prior steps, ensure that the conjugate base makes sense. This means that it should have one less hydrogen atom and it is one step more negative (or less positive) than the original acid. Validate this for each conjugate base obtained.

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Key Concepts

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

Conjugate Base Identification
The process of identifying a conjugate base involves analyzing how an acid behaves when it reacts with water. When an acid donates a proton (H+) to water, it becomes a conjugate base. This is because the acid loses a hydrogen ion.
  • The conjugate base is always associated with the acid that it arises from, known as its parent acid.
  • To find the conjugate base, look at the product side of the chemical equation. The substance that appears after the proton has been donated is the conjugate base.
  • For example, when reacting \( HIO_3 \) with water, it donates a proton to become \( IO_3^- \), the conjugate base.
A useful tip is to remember that a conjugate base will have one less hydrogen and usually amore negative charge compared to the original acid. Therefore, your task in chemical reaction problems is always simplified to identifying what remains after proton donation.
Chemical Equations
A chemical equation represents a chemical reaction where reactants are transformed into products. In the context of acid-base reactions, chemical equations help us visualize the transfer of protons. Writing these equations accurately is crucial for understanding the reaction process.
Here is a brief guide on how to write a chemical equation for acid reactions with water:
  • Start by writing the formula of the acid on the left side of the arrow, also known as the reactant.
  • Add water (usually written as \( H_2O \)) to the reactants, since it acts as the base accepting the proton.
  • Draw an arrow (↔) to separate the reactants from the products.
  • On the right side of the arrow, write the conjugate base, which is the acid minus one proton.
  • Finally, include \( H_3O^+ \) to show that water accepted the proton.
Each equation effectively captures the exchange of protons and helps us identify both the conjugate acid and base formed in these reactions. Make sure to balance the equation if necessary to ensure both mass and charge are conserved.
Proton Transfer
Proton transfer is a central event in acid-base chemistry. It distinguishes an acid-base reaction from other types of reactions. This mechanism involves the acid giving away a proton to the base. In an aqueous solution, water often plays the role of the base that accepts this proton.

Understanding proton transfer can be thought of as:
  • An acid is a proton donor, so it will lose a hydrogen ion during the reaction.
  • A base (like water) is a proton acceptor, therefore it will gain a hydrogen ion.
This transfer results in the formation of hydronium ions (\( H_3O^+ \)) and a conjugate base. The acidic strength is typically associated with how easily it gives up the proton, while the basic strength depends on its ability to accept it. Identifying the products of proton transfer allows us to understand the nature of acids and bases more thoroughly. This is crucial for predicting how different substances will react with each other in solution.

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Most popular questions from this chapter

Here is a way to test the validity of the statement made on page 719 in conjunction with the three key ideas governing the ionization of polyprotic acids. Determine the \(\mathrm{pH}\) of \(0.100 \mathrm{M}\) succinic acid in two ways: first by assuming that \(\mathrm{H}_{3} \mathrm{O}^{+}\) is produced only in the first ionization step, and then by allowing for the possibility that some \(\mathrm{H}_{3} \mathrm{O}^{+}\) is also produced in the second ionization step. Compare the results, and discuss the significance of your finding. $$\begin{array}{c} \mathrm{H}_{2} \mathrm{C}_{4} \mathrm{H}_{4} \mathrm{O}_{4}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{H}_{3} \mathrm{O}^{+}+\mathrm{HC}_{4} \mathrm{H}_{4} \mathrm{O}_{4}^{-} \\ &K_{\mathrm{a}_{1}}=6.2 \times 10^{-5} \\ \mathrm{HC}_{4} \mathrm{H}_{4} \mathrm{O}_{4}^{-}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{H}_{3} \mathrm{O}^{+}+\mathrm{C}_{4} \mathrm{H}_{4} \mathrm{O}_{4}^{2-} \\ &K_{\mathrm{a}_{2}}=2.3 \times 10^{-6} \end{array}$$

According to the Bronsted-Lowry theory, label each of the following as an acid or a base. (a) \(\mathrm{HNO}_{2}\) (b) \(\mathrm{OCl}^{-} ;(\mathrm{c}) \mathrm{NH}_{2}^{-} ;\) (d) \(\mathrm{NH}_{4}^{+} ;\) (e) \(\mathrm{CH}_{3} \mathrm{NH}_{3}^{+}\)

What is the (a) degree of ionization and (b) percent ionization of propionic acid in a solution that is \(0.45 \mathrm{M}\) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CO}_{2} \mathrm{H} ?\) $$\begin{aligned} &\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CO}_{2} \mathrm{H}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{H}_{3} \mathrm{O}^{+}+\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CO}_{2}^{-}\\\ &&\mathrm{p} K_{\mathrm{a}}=4.89 \end{aligned}$$

What are \(\left[\mathrm{H}_{3} \mathrm{O}^{+}\right],\left[\mathrm{OH}^{-}\right], \mathrm{pH},\) and \(\mathrm{pOH}\) of \(0.55 \mathrm{M}\) \(\mathrm{M} \mathrm{HClO}_{2} ?\)

Explain why \(\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]\) in a strong acid solution doubles as the total acid concentration doubles, whereas in a weak acid solution, \(\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]\) increases only by about a factor of \(\sqrt{2}\)

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