Problem 101 - Chemistry and Chemical Reactivity

Question

You prepare a 0.10 M solution of oxalic acid, \(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4} .\) What molecules and ions exist in this solution? List them in order of decreasing concentration.

Step-by-Step Solution

Answer

The molecules and ions are ordered as: \( H_2C_2O_4 > HC_2O_4^- > H^+ > C_2O_4^{2-} \).

Step 1: Understand the Dissociation Process

Oxalic acid, \(H_2C_2O_4\), is a diprotic acid, meaning it can lose two protons \(H^+\). Its dissociation can be presented in two steps. The first dissociation is \(nH_2C_2O_4 ightleftharpoons H^+ + HC_2O_4^-\), and the second dissociation \(nHC_2O_4^- ightleftharpoons H^+ + C_2O_4^{2-}\). In a 0.10 M solution, the first step occurs significantly more than the second, making \(H_2C_2O_4\) and \(HC_2O_4^-\) more prevalent.

Step 2: Identify Existing Species

In the 0.1 M solution of oxalic acid, you will find the following species due to dissociation:1. \(H_2C_2O_4\) (undissociated molecular form)2. \(HC_2O_4^-\) (hydrogen oxalate ion)3. \(H^+\) (proton ion)4. \(C_2O_4^{2-}\) (oxalate ion)Water \(H_2O\), the solvent, is also present but not altering the inherent concentrations of the solute.

Step 3: Arrange Species by Concentration

Based on the dissociation, the most concentrated species is the original acid \(H_2C_2O_4\), followed by the \(HC_2O_4^-\), as the first dissociation is partial. The \(H^+\) ions are next, as they accompany the first dissociation primarily, and \(C_2O_4^{2-}\) is the least because it requires the second dissociation.

Step 4: Arrange Species by Decreasing Concentration

1. \(H_2C_2O_4\)2. \(HC_2O_4^-\)3. \(H^+\)4. \(C_2O_4^{2-}\)This list reflects the order of concentration from highest to lowest.

Key Concepts

Diprotic Acid

A diprotic acid is an acid that can donate two protons or hydrogen ions (H^+) per molecule in a dissociation reaction. This characteristic is what distinguishes it from monoprotic acids, which only donate one proton. It is important because each of these protons can be donated in sequential reactions. When a diprotic acid like oxalic acid (H_2C_2O_4) is dissolved in water, it undergoes two separate ionization steps. The first ionization step involves the acid losing its first proton, resulting in the formation of the hydrogen oxalate ion (HC_2O_4^-). The second ionization step involves the hydrogen oxalate ion losing its second proton to form the oxalate ion (C_2O_4^{2-}).In equations, they are typically written as follows:

• First ionization step: \(H_2C_2O_4 \rightleftharpoons H^+ + HC_2O_4^- \)
• Second ionization step: \(HC_2O_4^- \rightleftharpoons H^+ + C_2O_4^{2-} \)

Understanding these steps is crucial, as the concentration of each species after dissolution depends on these reactions.

Oxalic Acid

Oxalic acid, with the chemical formula ( H_2C_2O_4 ), is a well-known diprotic acid. It is found in many plants and is known for its ability to form complexes with metal ions, which can make it both flavorful in certain foods and involved in biological processes. In aqueous solutions, oxalic acid undergoes dissociation in two stages:

• In the first stage, oxalic acid releases one proton to form HC_2O_4^- .
• In the second stage, the HC_2O_4^- ion releases another proton to form C_2O_4^{2-} .

When preparing a solution of oxalic acid, such as a 0.10 M solution, these dissociation steps become important in predicting the concentration of various species present in the solution. This behavior is what defines oxalic acid's chemical properties and reactivity.

Chemical Equilibrium

Chemical equilibrium is a state in which the concentrations of reactants and products remain constant over time. This happens when the rates of the forward and reverse reactions are equal. In the case of oxalic acid, the dissociation into its ions involves reversible reactions that establish equilibrium. As oxalic acid ( H_2C_2O_4 ) dissociates, it creates an equilibrium between the undissociated acid and its dissociation products ( HC_2O_4^- and C_2O_4^{2-} ). Key aspects of chemical equilibrium include:

• The equilibrium constant ( K_a ) for each ionization step indicates how far the reaction proceeds towards the products.
• At equilibrium, despite the ongoing interconversion of species, the quantities of each remain stable unless the system is disturbed.

Understanding chemical equilibrium helps predict which species will dominate in the solution and enables chemists to manipulate conditions to favor desired reactions.

Ionization Steps

Ionization steps refer to the sequential loss of protons in an acid, leading to the formation of different ions. For acids like oxalic acid, each proton is released in a separate step, making understanding these steps important in determining the behavior of the acid in solution. For oxalic acid:

• The first ionization step involves oxalic acid ( H_2C_2O_4 ) dissociating into a hydrogen ion ( H^+ ) and the hydrogen oxalate ion ( HC_2O_4^- ).
• The second ionization step involves the HC_2O_4^- ion losing another proton to form the oxalate ion ( C_2O_4^{2-} ).

These ionization steps do not happen to the same extent; the first ionization is usually more complete than the second, meaning more of the first ionization products (like HC_2O_4^- ) are usually present than the second ( C_2O_4^{2-} ). Using knowledge of ionization steps, we can predict and manipulate the chemistry of a solution.