Problem 15 - Chemistry and Chemical Reactivity

Question

Oxalic acid, \(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4},\) which is found in certain plants, can provide two hydrogen ions in water. Write balanced equations (like those for sulfuric acid on page 186 ) to show how oxalic acid can supply one and then a second \(\mathrm{H}^{+}\) ion.

Step-by-Step Solution

Answer

Oxalic acid releases two \(\mathrm{H}^{+}\) ions through two ionization steps.

Step 1: Write the Formula for Oxalic Acid

Oxalic acid is represented by the chemical formula \(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\). It is a diprotic acid, meaning it can donate two \(\mathrm{H}^{+}\) ions.

Step 2: Ionization of the First Hydrogen Ion

In the first ionization step, oxalic acid donates one \(\mathrm{H}^{+}\) ion to form the hydrogen oxalate ion. The chemical equation for this process is: \[\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}(aq) \rightarrow \mathrm{H}^{+}(aq) + \mathrm{HC}_{2} \mathrm{O}_{4}^{-}(aq)\]

Step 3: Ionization of the Second Hydrogen Ion

In the second ionization step, the hydrogen oxalate ion loses another \(\mathrm{H}^{+}\) ion to form the oxalate ion. The chemical equation for this step is: \[\mathrm{HC}_{2} \mathrm{O}_{4}^{-}(aq) \rightarrow \mathrm{H}^{+}(aq) + \mathrm{C}_{2} \mathrm{O}_{4}^{2-}(aq)\]

Step 4: Combined Balanced Equations

The combination of the two ionization steps gives the overall process: \[\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}(aq) \rightarrow 2\,\mathrm{H}^{+}(aq) + \mathrm{C}_{2} \mathrm{O}_{4}^{2-}(aq)\]

Key Concepts

Ionization

Ionization is a chemical process in which an atom or molecule gains a charge by losing or gaining electrons. In the context of acids, ionization refers to the process by which an acid releases hydrogen ions (H⁺) into a solution. Oxalic acid, like many acids, undergoes ionization when dissolved in water.
This process is essential because the release of hydrogen ions is what gives an acid its characteristic properties, such as sour taste and reactivity with metals.
The degree of ionization of an acid can affect its strength; stronger acids ionize more completely in water compared to weaker ones.

• One Step At A Time: Oxalic acid, being a diprotic acid, ionizes in two distinct steps.
• The first step involves losing one hydrogen ion, while the second involves losing another one from the resulting ion.
• These steps are crucial to understand, as each ionization step represents a different reaction that can influence the solution’s properties.

Balanced Chemical Equations

Balanced chemical equations are crucial for representing chemical reactions accurately. They show not only the substances involved but also the correct proportions in which these substances react.
This balance is achieved by ensuring that the number of atoms for each element is the same on both sides of the equation.
In the case of oxalic acid ionization, balancing helps to illustrate the step-by-step release of hydrogen ions.

• Step 1: The equation \[\mathrm{H}_{2}\mathrm{C}_{2}\mathrm{O}_{4}(aq) \rightarrow \mathrm{H}^{+}(aq) + \mathrm{HC}_{2}\mathrm{O}_{4}^{-}(aq)\] depicts the first ionization step, where one H⁺ ion and the hydrogen oxalate ion are formed, ensuring atom balance.
• Step 2: The subsequent equation \[\mathrm{HC}_{2} \mathrm{O}_{4}^{-}(aq) \rightarrow \mathrm{H}^{+}(aq) + \mathrm{C}_{2}\mathrm{O}_{4}^{2-}(aq)\] represents the second ionization step where another H⁺ ion forms along with the oxalate ion.
• Combining these ensures that every element's atoms remain balanced: the complete ionization shown in \[\mathrm{H}_{2}\mathrm{C}_{2}\mathrm{O}_{4}(aq) \rightarrow 2\mathrm{H}^{+}(aq) + \mathrm{C}_{2}\mathrm{O}_{4}^{2-}(aq)\].

Balancing these equations means every atom is accounted for, maintaining the law of conservation of mass.

Diprotic Acid

A diprotic acid is an acid that can donate two protons or hydrogen ions per molecule in an aqueous solution. This characteristic differentiates it from monoprotic acids, which can donate only one hydrogen ion.
Oxalic acid is a prime example of a diprotic acid.
Diprotic acids undergo ionization in two separate steps, each involving the release of one hydrogen ion.

• First Step: In the first ionization step, oxalic acid releases its first hydrogen ion, forming the hydrogen oxalate ion.
• Second Step: In the second ionization step, the hydrogen oxalate ion releases another hydrogen ion to form the oxalate ion.

In each step of ionization, the strength and concentration of ions in the solution can vary, affecting properties like pH. Understanding these steps is fundamental for predicting the behavior of diprotic acids in solutions and in chemical reactions.