Question

WEB Give the formula of the conjugate base of (a) \(\mathrm{HCO}_{3}^{-}\) (b) \(\mathrm{Cu}\left(\mathrm{H}_{2} \mathrm{O}\right)(\mathrm{OH})_{3}^{-}\) (c) \(\mathrm{HNO}_{2}\) (d) \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{NH}_{2}\) (e) \(\mathrm{H}_{2} \mathrm{SO}_{3}\)

Short answer

Question: Determine the conjugate base for each of the following substances: (a) \(\mathrm{HCO}_{3}^{-}\) (b) \(\mathrm{Cu}\left(\mathrm{H}_{2} \mathrm{O}\right)(\mathrm{OH})_{3}^{-}\) (c) \(\mathrm{HNO}_{2}\) (d) \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{NH}_{2}\) (e) \(\mathrm{H}_{2} \mathrm{SO}_{3}\) Answer: (a) \(\mathrm{CO}_{3}^{2-}\) (b) \(\mathrm{Cu}\left(\mathrm{H}_{2} \mathrm{O}\right)\left(\mathrm{OH}\right)_{2} \mathrm{O}^{2-}\) (c) \(\mathrm{NO}_{2}^{-}\) (d) \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{NH}^{-}\) (e) \(\mathrm{HSO}_{3}^{-}\)

Step-by-step solution

(a) Identifying the acidic hydrogen in \(\mathrm{HCO}_{3}^{-}\)

Observe that \(\mathrm{HCO}_{3}^{-}\) already has a net charge of -1, which indicates it has already lost a proton. The acidic hydrogen atom is the one attached to the oxygen atom.

(a) Writing the formula for conjugate base

After removing the acidic hydrogen (H⁺) from \(\mathrm{HCO}_{3}^{-}\), we are left with the formula \(\mathrm{CO}_{3}^{2-}\).

(b) Identifying the acidic hydrogen in \(\mathrm{Cu}\left(\mathrm{H}_{2} \mathrm{O}\right)(\mathrm{OH})_{3}^{-}\)

The acidic hydrogen is the one associated with the hydroxide (OH) groups.

(b) Writing the formula for conjugate base

Removing an acidic hydrogen (H⁺) from one of the hydroxide groups in \(\mathrm{Cu}\left(\mathrm{H}_{2} \mathrm{O}\right)(\mathrm{OH})_{3}^{-}\), we get the formula \(\mathrm{Cu}\left(\mathrm{H}_{2} \mathrm{O}\right)\left(\mathrm{OH}\right)_{2} \mathrm{O}^{2-}\).

(c) Identifying the acidic hydrogen in \(\mathrm{HNO}_{2}\)

The acidic hydrogen is the one attached to the oxygen atom in the nitrous acid (\(\mathrm{HNO}_{2}\)).

(c) Writing the formula for conjugate base

After removing the acidic hydrogen (H⁺) from \(\mathrm{HNO}_{2}\), we are left with the formula \(\mathrm{NO}_{2}^{-}\).

(d) Identifying the acidic hydrogen in \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{NH}_{2}\)

The acidic hydrogen is the one attached directly to the nitrogen atom.

(d) Writing the formula for conjugate base

After removing the acidic hydrogen (H⁺) from \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{NH}_{2}\), we are left with the formula \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{NH}^{-}\).

(e) Identifying the acidic hydrogen in \(\mathrm{H}_{2} \mathrm{SO}_{3}\)

The acidic hydrogens are the ones attached to the oxygen atoms in sulfurous acid (\(\mathrm{H}_{2} \mathrm{SO}_{3}\)).

(e) Writing the formula for conjugate base

After removing an acidic hydrogen (H⁺) from \(\mathrm{H}_{2} \mathrm{SO}_{3}\), we are left with the formula \(\mathrm{HSO}_{3}^{-}\).

Key concepts

Acidic Hydrogen

Acidic hydrogen atoms are special types of hydrogen atoms in a molecule, as they can be easily removed. This makes them critical in determining the acidity of a compound. These hydrogen atoms are typically bonded to more electronegative atoms such as oxygen or nitrogen, which makes them potentially more acidic.
When we say a hydrogen is acidic, it means that the hydrogen can be liberated as a proton (H⁺). In essence, the bond between the hydrogen and the adjacent atom is sufficiently polarized due to the difference in electronegativity.
For example, in the bicarbonate ion (\(\mathrm{HCO}_{3}^{-}\)), the acidic hydrogen is bonded to an oxygen atom. This bond is polarized, making the hydrogen slightly positive and prone to release, making the ion acidic.

Chemical Formula

A chemical formula represents the types and numbers of atoms in a molecule. For students trying to identify conjugate bases, understanding chemical formulas is crucial. Each element in the formula is denoted by its chemical symbol, such as H for hydrogen or O for oxygen.
It also includes subscripts to indicate the number of each type of atom in the molecule. For instance, in \(\mathrm{Cu(H_{2}O)(OH)_{3}^{-}\)), the subscript tells us there is one copper (Cu), two hydrogen atoms in water (H₂O), and three hydroxide ions (OH⁻). Formulas also convey charge, like the -1 charge on this compound, indicating it has gained an extra electron.
Knowing how to read and interpret chemical formulas helps in predicting the behavior of the molecule in chemical reactions, such as the loss of an acidic hydrogen atom.

Proton Removal

The concept of proton removal is central to understanding acids and bases. When a molecule loses an acidic hydrogen, it effectively loses a proton (\(\mathrm{H}^+\)), because a hydrogen atom is essentially a proton with an electron.
This removal forms the conjugate base of the acid, which is a key concept in Brønsted-Lowry theory. For instance, when removing a proton from \(\mathrm{HNO}_{2}\) (nitrous acid), we end up with \(\mathrm{NO}_{2}^{-}\), its conjugate base.
It's essential to understand that the removal of a proton is a reversible process, highlighting the dynamic nature of acids and bases in equilibrium.

Conjugate Acid-Base Pair

In acid-base chemistry, a conjugate acid-base pair consists of two species that differ by just one hydrogen ion (\(\mathrm{H}^+\)). The acid of the pair donates a proton, while the base accepts a proton.
This pair exists in equilibrium, as one can convert to the other by either gaining or losing a proton. Taking \(\mathrm{H}_{2}\mathrm{SO}_{3}\) (sulfurous acid) as an example, when it donates an acidic hydrogen, it forms \(\mathrm{HSO}_{3}^{-}\), its conjugate base.
Understanding this relationship helps in predicting the direction of acid-base reactions and in forming a deeper understanding of chemical equilibrium and acidity.