Question

The carbonate ion, \(\mathrm{CO}_{3}^{2}\), is a weak base. (a) Write the equilibrium equation that shows how this ion makes water basic. (b) Which species is the acid and which species is the base according to the Brønsted-Lowry definition? (c) The equilibrium constant for the reaction between \(\mathrm{CO}_{3}^{2-}\) and water is \(2.1 \times 10^{-4}\) What does this tell you about the strength of Che 2 a base?

Short answer

(a) The equilibrium equation for the carbonate ion making water basic is: \[\mathrm{CO}_{3}^{2-} + \mathrm{H}_{2}\mathrm{O} \rightleftharpoons \mathrm{OH}^- + \mathrm{HCO}_{3}^-\] (b) According to the Brønsted-Lowry definition, in this reaction: 1. The carbonate ion (\(\mathrm{CO}_{3}^{2-}\)) is the base as it accepts a proton. 2. Water (\(\mathrm{H}_{2}\mathrm{O}\)) is the acid as it donates a proton. (c) The equilibrium constant for the reaction between the carbonate ion and water is given as \(2.1 \times 10^{-4}\). This small value confirms that the \(\mathrm{CO}_{3}^{2-}\) ion is a weak base.

Step-by-step solution

(a) Writing the equilibrium equation for the carbonate ion

The carbonate ion (\(\mathrm{CO}_{3}^{2-}\)) reacts with water (\(\mathrm{H}_{2}\mathrm{O}\)) to form a hydroxide ion (\(\mathrm{OH}^-\)) and a bicarbonate ion (\(\mathrm{HCO}_{3}^-\)). The equilibrium equation is: \[\mathrm{CO}_{3}^{2-} + \mathrm{H}_{2}\mathrm{O} \rightleftharpoons \mathrm{OH}^- + \mathrm{HCO}_{3}^-\]

(b) Identifying Brønsted-Lowry acid and base species

According to the Brønsted-Lowry definition, acids donate a proton (\(\mathrm{H}^+\)) and bases accept a proton. In this reaction: 1. The carbonate ion (\(\mathrm{CO}_{3}^{2-}\)) accepts a proton and is thus the Brønsted-Lowry base. 2. Water (\(\mathrm{H}_{2}\mathrm{O}\)) donates a proton, forming a hydroxide ion (\(\mathrm{OH}^-\)) and is the Brønsted-Lowry acid.

(c) Interpreting the equilibrium constant and the strength of the base

The equilibrium constant for the reaction between the carbonate ion and water is given as \(2.1 \times 10^{-4}\). The equilibrium constant (\(K_{b}\)) is used to describe the relative strength of a base. A larger \(K_{b}\) value corresponds to a stronger base, while a smaller value indicates a weaker base. Since the equilibrium constant for this reaction is small (\(2.1 \times 10^{-4}\)), it confirms that the \(\mathrm{CO}_{3}^{2-}\) ion is a weak base.

Key concepts

Brønsted-Lowry acid and base

In chemistry, the Brønsted-Lowry theory is a widely used concept to describe acids and bases. According to this theory, an acid is a substance that can donate a proton (\(\mathrm{H}^+\)), while a base is a substance that can accept a proton. This definition helps us understand many chemical reactions, especially those involving aqueous solutions.
In the context of the carbonate ion (\(\mathrm{CO}_{3}^{2-}\)) reacting with water (\(\mathrm{H}_{2}\mathrm{O}\)), this concept is crucial. Here, the water molecule acts as a Brønsted-Lowry acid because it donates a proton to the carbonate ion. The carbonate ion, by accepting this proton, is considered a Brønsted-Lowry base.

• In the given reaction:\(\phantom{abc}\mathrm{CO}_{3}^{2-} + \mathrm{H}_{2}\mathrm{O} \rightleftharpoons \mathrm{OH}^- + \mathrm{HCO}_{3}^-\)
• The carbonate ion (\(\mathrm{CO}_{3}^{2-}\)) is the base, accepting a proton to form bicarbonate (\(\mathrm{HCO}_{3}^-\)).
• Water (\(\mathrm{H}_{2}\mathrm{O}\)) is the acid, donating a proton to form a hydroxide ion (\(\mathrm{OH}^-\)).

Understanding the Brønsted-Lowry acid-base concept allows us to predict and explain the behavior of substances in various chemical reactions.

Equilibrium Constant

The equilibrium constant (\(K_{b}\)) in a chemical reaction is a crucial value that helps determine the strength and direction of the reaction. It tells us how much the reactants (on the left side of the equilibrium equation) are favored compared to the products (on the right side).
For the carbonate ion reaction\(\mathrm{CO}_{3}^{2-} + \mathrm{H}_{2}\mathrm{O} \rightleftharpoons \mathrm{OH}^- + \mathrm{HCO}_{3}^-\), the equilibrium constant is given as\(K_{b} = 2.1 \times 10^{-4}\). This value is derived from the concentrations of the products and reactants at equilibrium.

• A small equilibrium constant, as in this case, means that the formation of products (bicarbonate and hydroxide) is not heavily favored, indicating that the carbonate ion does not fully convert to its products.
• Therefore, the reaction strongly favors the reactants (\(\mathrm{CO}_{3}^{2-}\) and\(\mathrm{H}_{2}\mathrm{O}\)).

Thus, the equilibrium constant gives us insight into the reaction dynamics, showing the extent to which the carbonate ion can make the solution basic.

Weak Base

A weak base is a substance that does not fully dissociate in water, meaning it only partially accepts protons from water molecules. The carbonate ion (\(\mathrm{CO}_{3}^{2-}\)) is classified as a weak base because it partially reacts with water to produce hydroxide ions (\(\mathrm{OH}^-\)) and bicarbonate ions (\(\mathrm{HCO}_{3}^-\)).
The equilibrium constant (\(K_{b}\)) of\(2.1 \times 10^{-4}\) reinforces this classification. A small\(K_{b}\) value indicates that only a minor portion of the carbonate ions convert to hydroxide and bicarbonate, leaving most of them in their original state. This behavior is typical for weak bases:

• They do not react completely with water; instead, an equilibrium is established.
• Their solutions have a higher pH than neutral water but are less basic than those of strong bases.

Understanding weak bases is essential for predicting the behavior and pH of solutions in which they are dissolved, leading to applications in fields ranging from environmental chemistry to biochemistry.