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Chapter 10: Problem 48

Write the formulas of the conjugate acids of the following Br酶nsted-Lowry bases: (a) \(\mathrm{ClCH}_{2} \mathrm{CO}_{2}^{-}\) (b) \(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{~N}\) (c) \(\mathrm{SeO}_{4}^{2-}\) (d) \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{~N}\)

Short Answer

Expert verified
(a) ClCH鈧侰OOH (b) C鈧匟鈧匩H鈦� (c) HSeO鈧勨伝 (d) (CH鈧�)鈧僋H鈦�

Step by step solution

01

Understanding Conjugate Acids

To find the conjugate acid of a Br酶nsted-Lowry base, we need to add a proton (H鈦�) to the base. The addition of a proton converts the base into its conjugate acid form.
02

Find the Conjugate Acid of ClCH鈧侰O鈧傗伝

For the base \( \mathrm{ClCH}_{2} \mathrm{CO}_{2}^{-} \), add a proton \( \left( \mathrm{H}^{+} \right) \). This results in \( \mathrm{ClCH}_{2} \mathrm{CO}_{2H} \), also known as \( \mathrm{ClCH}_{2} \mathrm{COOH} \).
03

Find the Conjugate Acid of C鈧匟鈧匩

For the base \( \mathrm{C}_{5} \mathrm{H}_{5} \mathrm{~N} \), add a proton \( \left( \mathrm{H}^{+} \right) \). This results in \( \mathrm{C}_{5} \mathrm{H}_{5} \mathrm{NH}^{+} \).
04

Find the Conjugate Acid of SeO鈧劼测伝

For the base \( \mathrm{SeO}_{4}^{2-} \), add a proton \( \left( \mathrm{H}^{+} \right) \). This results in \( \mathrm{HSeO}_{4}^{-} \).
05

Find the Conjugate Acid of (CH鈧�)鈧僋

For the base \( \left( \mathrm{CH}_{3} \right)_{3} \mathrm{~N} \), add a proton \( \left( \mathrm{H}^{+} \right) \). This results in \( \left( \mathrm{CH}_{3} \right)_{3} \mathrm{NH}^{+} \).

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Key Concepts

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

Conjugate Acid
A conjugate acid forms when a Br酶nsted-Lowry base gains a proton. This process transforms the base molecule into a new species, the conjugate acid. Simply speaking, when a base accepts a hydrogen ion (ul>
  • It results in a new chemical entity.
  • The formula changes to include one more hydrogen atom.
  • A positive charge may appear due to the gained proton.
    • For example, if we start with the base \( \mathrm{ClCH}_{2} \mathrm{CO}_{2}^{-} \), adding a proton gives us \( \mathrm{ClCH}_{2} \mathrm{COOH} \), indicating this process.Being able to identify the conjugate acid allows us to understand how substances can act as buffers or participate in various chemical processes.
    Protonation
    Protonation refers to the process where a proton (H\(^{+}\)) is added to a molecule. It is a key step in forming conjugate acids from bases. The gain of this proton changes the base to its conjugate acid.
    • This simple addition is crucial in many acid-base reactions.
    • Understanding protonation provides insight into the reactivity of compounds.
    Let鈥檚 take the base \( \mathrm{C}_{5} \mathrm{H}_{5} \mathrm{~N} \) as an example. Through protonation, a proton is added, resulting in \( \mathrm{C}_{5} \mathrm{H}_{5} \mathrm{NH}^{+} \). This reveals not only the transforming effect a proton can have but also the dynamic nature of chemical reactions.
    Chemical Formulas
    Chemical formulas represent the composition of molecules. They show the elements involved and their quantities. In the context of finding conjugate acids, chemical formulas help us understand the molecular changes during protonation.
    • Changes in chemical formulas can signal protonation.
    • Notation like +(an additional element or charge) denotes new elements or charges added.
    Consider the base \( \mathrm{SeO}_{4}^{2-} \). Ionically, this base gains a proton, becoming \( \mathrm{HSeO}_{4}^{-} \). Understanding the chemical formulas is key to grasping how compounds react and transform.
    Step-by-Step Solution
    A step-by-step solution involves breaking down complex processes into simpler, manageable parts. This approach helps tackle chemical exercises, like determining conjugate acids, more effectively.
    • It allows a structured way to solve chemical problems.
    • Encourages working through problems methodically.
    For example, if we have a base like \( (\mathrm{CH}_{3})_{3} \mathrm{~N} \), the steps are: 1. Identify the base. 2. Add a proton. 3. Write the new formula: \( (\mathrm{CH}_{3})_{3} \mathrm{NH}^{+} \). This helps in tracking changes and understanding the chemical behavior. Through these steps, questions become more approachable and solvable.

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    Most popular questions from this chapter

    The hydrogen-containing anions of many polyprotic acids are amphoteric. Write equations for \(\mathrm{HCO}_{3}^{-}\) and \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\) acting as bases with the strong acid \(\mathrm{HCl}\) and as acids with the strong base \(\mathrm{NaOH}\).

    Which solution would have the higher \(\mathrm{pH}: 0.010 \mathrm{M} \mathrm{HNO}_{2}\) or \(0.010 \mathrm{M} \mathrm{HNO}_{3}\) ? Explain.

    What are the molarity and the normality of a solution made by dissolving \(25 \mathrm{~g}\) of citric acid (triprotic, \(\left.\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{O}_{7} \mathrm{H}_{3}\right)\) in enough water to make \(800 \mathrm{~mL}\) of solution?

    Many allergy medications contain antihistamines, compounds that contain amine groups \(\left(\mathrm{R}-\mathrm{NH}_{2},\right.\) where \(\mathrm{R}\) refers to an organic functional group). Would you expect these compounds to be acidic, basic or neutral? Explain. (a) One over-the-counter product lists the active ingredient as "diphenhydramine HCl." What does this designation mean? (b) Write the acid-base reaction to illustrate how this compound is produced. When this product is dissolved in water would you expect the solution be acidic, basic, or neutral? Explain.

    Label the Br酶nsted-Lowry acids and bases in the following equations, and tell which substances are conjugate acid-base pairs. $$ \begin{array}{l} \text { (a) } \mathrm{CO}_{3}^{2-}(a q)+\mathrm{HCl}(a q) \longrightarrow \mathrm{HCO}_{3}^{-}(a q)+\mathrm{Cl}^{-}(a q) \\ \text { (b) } \mathrm{H}_{3} \mathrm{PO}_{4}(a q)+\mathrm{NH}_{3}(a q) \longrightarrow \\ & \mathrm{H}_{2} \mathrm{PO}_{4}^{-}(a q)+\mathrm{NH}_{4}^{+}(a q) \\ \text { (c) } \mathrm{NH}_{4}^{+}(a q)+\mathrm{CN}^{-}(a q) \rightleftarrows \mathrm{NH}_{3}(a q)+\mathrm{HCN}(a q) \\ \text { (d) } \mathrm{HBr}(a q)+\mathrm{OH}^{-}(a q) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{Br}^{-}(a q) \end{array} $$ (e) \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}(a q)+\mathrm{N}_{2} \mathrm{H}_{4}(a q) \rightleftarrows\) $$ \mathrm{HPO}_{4}^{2-}(a q)+\mathrm{N}_{2} \mathrm{H}_{5}^{+}(a q) $$
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