Problem 13 - Chemistry Matter and Change

Question

Write the first and second ionization equations for \(\mathrm{H}_{2} \mathrm{SeO}_{3}\)

Step-by-Step Solution

Answer

The first and second ionization equations for \(\mathrm{H}_{2}\mathrm{SeO}_{3}\) are as follows: 1. \(\mathrm{H}_{2}\mathrm{SeO}_{3} \rightarrow \mathrm{H}_{1}\mathrm{SeO}_{3}^{-} + \mathrm{H}^{+}\) 2. \(\mathrm{H}_{1}\mathrm{SeO}_{3}^{-} \rightarrow \mathrm{SeO}_{3}^{2-} + \mathrm{H}^{+}\)

Step 1: Identify the ions produced in the first ionization

For the first ionization, \(\mathrm{H}_{2}\mathrm{SeO}_{3}\) donates one proton (H+) and forms a new substance, thus producing one ion. This newly formed ion will have one less hydrogen.

Step 2: Write the first ionization equation

The first ionization equation will be: \[\mathrm{H}_{2}\mathrm{SeO}_{3} \rightarrow \mathrm{H}_{1}\mathrm{SeO}_{3}^{-} + \mathrm{H}^{+}\] Step 2: Write the second ionization equation

Step 3: Identify the ions produced in the second ionization

Now, the product that results from the first ionization, \(\mathrm{H}_{1}\mathrm{SeO}_{3}^{-}\), can donate another proton (H+) and form a new substance, thus producing another ion. This newly formed ion will have no hydrogens left.

Step 4: Write the second ionization equation

The second ionization equation will be: \[\mathrm{H}_{1}\mathrm{SeO}_{3}^{-} \rightarrow \mathrm{SeO}_{3}^{2-} + \mathrm{H}^{+}\] The first and second ionization equations for \(\mathrm{H}_{2}\mathrm{SeO}_{3}\) are then: 1. \(\mathrm{H}_{2}\mathrm{SeO}_{3} \rightarrow \mathrm{H}_{1}\mathrm{SeO}_{3}^{-} + \mathrm{H}^{+}\) 2. \(\mathrm{H}_{1}\mathrm{SeO}_{3}^{-} \rightarrow \mathrm{SeO}_{3}^{2-} + \mathrm{H}^{+}\)

Key Concepts

Chemical Equations

A chemical equation is a written representation of a chemical reaction. It uses symbols and formulas to display the substances involved. On the left, you write the reactants—substances that start the reaction. On the right, you'll find the products—substances formed by the reaction.

In chemistry, it's important to balance these equations to obey the law of conservation of mass. This law states that the mass and number of atoms do not change during a chemical reaction. Therefore, the number of atoms for each element must be equal on both sides of the equation. For instance, in the ionization of \(\mathrm{H}_2\mathrm{SeO}_3\), the first equation is balanced as it reflects the correct number of hydrogens, selenium, and oxygen molecules.

To elaborate, \(\mathrm{H}_2\mathrm{SeO}_3\) donates a proton to become \(\mathrm{H}_1\mathrm{SeO}_3^{-}\), and then it can again donate a proton to form \(\mathrm{SeO}_3^{2-}\). Each step reflects a balanced chemical equation in terms of mass and charge.

Ionization Process

The ionization process involves turning atoms or molecules into ions. This occurs by adding or removing charged particles, such as electrons or protons. Specifically for acids, this usually means donating protons (\(\mathrm{H}^+\)).

In cases like \(\mathrm{H}_2\mathrm{SeO}_3\), which is a diprotic acid, ionization occurs in steps. The first ionization removes one \(\mathrm{H}^+\), forming \(\mathrm{H}_1\mathrm{SeO}_3^{-}\). The second ionization removes another \(\mathrm{H}^+\), resulting in \(\mathrm{SeO}_3^{2-}\). Each ionization step results in a negatively charged ion and requires the input of energy to remove the proton.

This stepwise removal is typical of polyprotic acids, which can donate more than one proton. Each step typically becomes progressively harder and needs more energy, as removing a proton from an already charged ion is more challenging.

Proton Donation

Proton donation is a fundamental concept in acid-base chemistry. It involves an acid releasing \(\mathrm{H}^+\) ions into a solution. In the water, these ions are often referred to as hydronium ions (\(\mathrm{H}_3\mathrm{O}^+\)).

For \(\mathrm{H}_2\mathrm{SeO}_3\), proton donation occurs in two sequential steps, exemplifying its nature as a diprotic acid. In the first step, \(\mathrm{H}_2\mathrm{SeO}_3\) donates a proton, forming \(\mathrm{H}_1\mathrm{SeO}_3^{-}\). This is the first stage of its chemical reactivity.

The second proton donation occurs from \(\mathrm{H}_1\mathrm{SeO}_3^{-}\), transforming it into \(\mathrm{SeO}_3^{2-}\). Each proton that is donated is a key part of the molecule’s ability to create ions and participate in further chemical reactions. The strength and effectiveness of acids often depend on how willing they are to donate protons.