Problem 2 - Chemical Principles in the Laboratory

Question

A solution of \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) has a \(\mathrm{pH}\) of \(10 .\) The \(\mathrm{CO}_{3}^{2-}\) ion is the conjugate base of the \(\mathrm{HCO}_{3}^{-}\) ion. Write the net ionic equation for the reaction which makes a solution of \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) basic. (Eq. \(7 .\) )

Step-by-Step Solution

Answer

The net ionic equation for the reaction that makes a solution of \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) basic is: \[ \mathrm{CO}_{3}^{2-} + \mathrm{H}_{2}\mathrm{O} \leftrightarrow \mathrm{HCO}_{3}^{-} + \mathrm{OH}^{-} \]

Step 1: Dissolution of \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) in water

When \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) is dissolved in water, it dissociates into its constituent ions as follows: \[ \mathrm{Na}_{2} \mathrm{CO}_{3} \rightarrow 2 \mathrm{Na}^{+} + \mathrm{CO}_{3}^{2-} \]

Step 2: Reaction of \(\mathrm{CO}_{3}^{2-}\) (conjugate base) with water

Since \(\mathrm{CO}_{3}^{2-}\) ion is the conjugate base of the \(\mathrm{HCO}_{3}^{-}\) ion, it reacts with water to receive a proton (H\(^{+}\)) to form \(\mathrm{HCO}_{3}^{-}\). The reaction is as follows: \[ \mathrm{CO}_{3}^{2-} + \mathrm{H}_{2}\mathrm{O} \leftrightarrow \mathrm{HCO}_{3}^{-} + \mathrm{OH}^{-} \]

Step 3: Write the net ionic equation

The net ionic equation involves only the ions that participate in the reaction, which are \(\mathrm{CO}_{3}^{2-}\), \(\mathrm{H}_{2}\mathrm{O}\), \(\mathrm{HCO}_{3}^{-}\), and \(\mathrm{OH}^{-}\). Thus, the net ionic equation for the reaction that makes the solution of \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) basic is: \[ \mathrm{CO}_{3}^{2-} + \mathrm{H}_{2}\mathrm{O} \leftrightarrow \mathrm{HCO}_{3}^{-} + \mathrm{OH}^{-} \]

Key Concepts

Chemical Equilibrium

When we talk about chemical equilibrium, we're discussing a state where the rate of the forward reaction is equal to the rate of the reverse reaction in a chemical process. This balance means that the concentrations of the reactants and products remain constant over time, not changing even though both reactions continue to occur.

In the context of the exercise, when sodium carbonate (Na_2CO_3) is dissolved in water, the carbonate ion (CO_3^{2-}) reaches a state of equilibrium with bicarbonate (HCO_3^{-}) and hydroxide ions (OH^{-}). The equation given in the solution, CO_3^{2-} + H_2O leftrightarrow HCO_3^{-} + OH^{-}, represents this equilibrium. Over time, the amounts of these ions in the solution won't change, indicating a dynamic equilibrium has been achieved—as many carbonate ions are grabbing a proton to become bicarbonate as bicarbonate ions are releasing a proton to revert to carbonate.

Acid-Base Conjugate Pairs

The concept of acid-base conjugate pairs is central to understanding many reactions that occur in aqueous solutions, including the one in our given exercise. An acid is a substance that can donate a proton (H^{+}), while a base is a substance that can accept a proton. In an acid-base reaction, the acid donates a proton to the base, forming a conjugate base and a conjugate acid.

The carbonate ion (CO_3^{2-}), which is the focus of our exercise, acts as the base and accepts a proton from water, which is acting as an acid, to become bicarbonate (HCO_3^{-}). H_2O, after giving up a proton, becomes the conjugate base (OH^{-}). The original base (CO_3^{2-}) and its corresponding conjugate acid (HCO_3^{-}) form one conjugate pair, while water (H_2O) and hydroxide ion (OH^{-}) form another conjugate pair. This transfer of protons is the hallmark of acid-base chemistry and is illustrated in the net ionic equation provided.

Dissociation of Ionic Compounds

Understanding the dissociation of ionic compounds is pivotal for grasping the behavior of salts in solution, such as the dissolution of sodium carbonate (Na_2CO_3) in our exercise. Dissociation refers to the process by which an ionic compound separates into its positive and negative ions when it dissolves in water. This separation happens because water molecules are polar, meaning they have a partial positive charge near the hydrogen atoms and a partial negative charge near the oxygen atom.

When Na_2CO_3 dissolves, the polar water molecules surround and stabilize the sodium (Na^+) and carbonate (CO_3^{2-}) ions, pulling them apart from the solid lattice structure in which they are held together. The ability of water to do this makes it a very effective solvent for ionic compounds. In the solution presented, the first step clearly shows the dissociation reaction, where Na_2CO_3 splits into two sodium ions (2 Na^+) and one carbonate ion (CO_3^{2-}). This process is essential for the subsequent reactions that lead to the basic nature of the solution.