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Chapter 15: Problem 12

The bicarbonate ion, \(\mathrm{HCO}_{3}^{-}\), has the ability to act as both an acid and a base. (a) What is the conjugate acid of \(\mathrm{HCO}_{3}^{-}\)? (b) What is the conjugate base of \(\mathrm{HCO}_{3}^{-}\)?

Short Answer

Expert verified
(a) \(\text{H}_2\text{CO}_3\); (b) \(\text{CO}_{3}^{2-}\).

Step by step solution

01

Identify the conjugate acid

To determine the conjugate acid of \(\text{HCO}_{3}^{-}\), add a \(\text{H}^+\) ion to the bicarbonate ion. The addition of one proton (\(\text{H}^+\)) to \(\text{HCO}_{3}^{-}\) results in the formation of carbonic acid, \(\text{H}_2\text{CO}_3\). Therefore, the conjugate acid of \(\text{HCO}_{3}^{-}\) is \(\text{H}_2\text{CO}_3\).
02

Identify the conjugate base

To find the conjugate base of \(\text{HCO}_{3}^{-}\), remove a \(\text{H}^+\) ion from the bicarbonate ion. The removal of one proton (\(\text{H}^+\)) from \(\text{HCO}_{3}^{-}\) results in the formation of the carbonate ion, \(\text{CO}_{3}^{2-}\). Therefore, the conjugate base of \(\text{HCO}_{3}^{-}\) is \(\text{CO}_{3}^{2-}\).

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Key Concepts

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

Conjugate Acid
In acid-base chemistry, a conjugate acid is formed when a base accepts a proton \((\text{H}^+)\). For example, when the bicarbonate ion \( \text{HCO}_3^- \) acts as a base, it can accept a proton to become carbonic acid \( \text{H}_2\text{CO}_3 \). This process is described by the equation:\
\[ \text{HCO}_3^- + \text{H}^+ \rightarrow \text{H}_2\text{CO}_3 \] \ The bicarbonate ion \( \text{HCO}_3^- \) gains a hydrogen ion and becomes its conjugate acid, carbonic acid.
Conjugate Base
A conjugate base is what is left after an acid donates a proton \((\text{H}^+)\). For instance, when the bicarbonate ion \( \text{HCO}_3^- \) acts as an acid, it can donate a proton to form the carbonate ion \( \text{CO}_3^{2-} \). This can be seen in the equation below:
\[ \text{HCO}_3^- \rightarrow \text{CO}_3^{2-} + \text{H}^+ \] \ After \( \text{HCO}_3^- \) loses a hydrogen ion, it becomes its conjugate base \ ( \text{CO}_3^{2-})\.
Acid-Base Chemistry
Acid-base chemistry revolves around the concepts of acids and bases, which are substances that can donate or accept protons, respectively. When an acid donates a proton, it forms its conjugate base. Conversely, when a base accepts a proton, it forms its conjugate acid.
  • An acid is a proton donor.
  • A base is a proton acceptor.
  • A conjugate acid is the substance formed when a base gains a proton.
  • A conjugate base is the substance formed when an acid loses a proton.
The bicarbonate ion \( \text{HCO}_3^- \) is special because it can behave as both an acid and a base. This dual behavior is known as 'amphoterism'. Here's a quick recap:
  • As a base: \[ \text{HCO}_3^- + \text{H}^+ \rightarrow \text{H}_2\text{CO}_3 \]
  • As an acid: \[ \text{HCO}_3^- \rightarrow \text{CO}_3^{2-} + \text{H}^+ \]
Understanding these fundamental principles is key to mastering acid-base reactions and identifying conjugate acids and bases.

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Most popular questions from this chapter

Given the data for the following separate titrations, calculate the molarity of the \(\mathrm{HCl}\) : $$ \begin{array}{|c|c|c|c|c|} \hline & \begin{array}{c} \text { mL } \\ \text { HCl } \end{array} & \begin{array}{c} \text { Molarity } \\ \text { HCl } \end{array} & \begin{array}{c} \mathbf{m L} \\ \mathbf{N a O H} \end{array} & \begin{array}{c} \text { Molarity } \\ \mathbf{N a O H} \end{array} \\ \hline \text { (a) } & 40.13 & M & 37.70 & 0.728 \\ \text { (b) } & 19.00 & M & 33.66 & 0.306 \\ \text { (c) } & 27.25 & M & 18.00 & 0.555 \\ \hline \end{array} $$

Determine how many milliliters of \(0.525 \mathrm{M} \mathrm{HClO}_{4}\) will be required to neutralize \(50.25 \mathrm{~g} \mathrm{Ca}(\mathrm{OH})_{2}\) according to the reaction: $$ 2 \mathrm{HClO}_{4}(a q)+\mathrm{Ca}(\mathrm{OH})_{2}(s) \longrightarrow \mathrm{Ca}\left(\mathrm{ClO}_{4}\right)_{2}(a q)+2 \mathrm{H}_{2} \mathrm{O}(l) $$

Determine whether each of the following describes a substance that is acidic, basic, or neutral: (a) Red litmus paper turns blue (b) \(\mathrm{pH}=6\) (c) \(\left[\mathrm{H}^{+}\right]=1 \times 10^{-7}\) (d) \(\left[\mathrm{OH}^{-}\right]=1 \times 10^{-8}\) (e) Blue litmus paper turns red (f) \(\mathrm{pH}=12\)

Determine how many grams of \(\mathrm{Al}(\mathrm{OH})_{3}\) will be required to neutralize \(275 \mathrm{~mL}\) of \(0.125 \mathrm{M} \mathrm{HCl}\) according to the reaction: $$ 3 \mathrm{HCl}(a q)+\mathrm{Al}(\mathrm{OH})_{3}(s) \longrightarrow \mathrm{AlCl}_{3}(a q)+3 \mathrm{H}_{2} \mathrm{O}(l) $$

A \(0.200\)-g sample of impure \(\mathrm{NaOH}\) requires \(18.25 \mathrm{~mL}\) of \(0.2406\) \(M \mathrm{HCl}\) for neutralization. What is the percent of \(\mathrm{NaOH}\) in the sample?
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