Problem 21 - Chemistry: An Atoms-Focused Approach

Question

What is the concentration of \(\mathrm{H}_{3} \mathrm{O}^{+}\) ions in \(0.65 \mathrm{M} \mathrm{HNO}_{3} ?\)

Step-by-Step Solution

Answer

Answer: The concentration of \(\mathrm{H}_{3}\mathrm{O}^{+}\) ions in the 0.65 M solution of HNO\(_3\) is 0.65 M.

Step 1: Write the chemical equation for HNO\(_3\) ionization

To find the concentration of \(\mathrm{H}_{3}\mathrm{O}^{+}\) ions in a solution of HNO\(_3\), we first need to write the chemical equation for its ionization in the water. The ionization of HNO\(_3\) can be given as: HNO\(_3\) + H\(_2\)O → \(\mathrm{H}_{3}\mathrm{O}^{+}\) + NO\(_3^-\)

Step 2: Analyze the stoichiometry of the balanced equation

From the balanced equation in Step 1, we can see that one mole of HNO\(_{3}\) will produce one mole of \(\mathrm{H}_{3}\mathrm{O}^{+}\) ions. The stoichiometry of HNO\(_{3}\) and \(\mathrm{H}_{3}\mathrm{O}^{+}\) is a 1:1 ratio.

Step 3: Calculate the concentration of \(\mathrm{H}_{3}\mathrm{O}^{+}\) ions

Since HNO\(_3\) is a strong acid, it will completely ionize in water. Therefore, the concentration of \(\mathrm{H}_{3}\mathrm{O}^{+}\) ions produced will be equal to the concentration of HNO\(_3\). So, in this case, the concentration of HNO\(_3\) provided in the exercise is 0.65 M, and the concentration of \(\mathrm{H}_{3}\mathrm{O}^{+}\) ions will be the same: \(\mathrm{H}_{3}\mathrm{O}^{+}\ \text{concentration} = 0.65\ \text{M}\) Now we have the concentration of the \(\mathrm{H}_{3}\mathrm{O}^{+}\) ions in the 0.65 M solution of HNO\(_{3}\).

Step 4: Present the final answer

The concentration of \(\mathrm{H}_{3}\mathrm{O}^{+}\) ions in a 0.65 M solution of HNO\(_3\) is 0.65 M.

Key Concepts

Concentration Calculations

When working with solutions, particularly those involving acids and bases, understanding concentration calculations is crucial. Concentration tells us how much of a substance is present in a certain amount of a solution, typically expressed in molarity (M), which is moles per liter.
The concentration of ions in a solution can often be found through simple stoichiometry, especially when dealing with a strong acid or base. In the original problem, we were asked to find the concentration of hydronium ions, \(\mathrm{H}_{3}\mathrm{O}^{+}\), which mirrors the concentration of HNO\(_3\) due to its complete ionization in water. This is key because many acids or bases may not completely ionize.
Thus, steps to solve concentration problems include:

• Understanding the chemical equation involved.
• Noting stoichiometry and molar relationships.
• Recognizing whether the compound is a strong or weak electrolyte.

These considerations make the concentration calculation straightforward and ensure accuracy in your results.

Strong Acids

Strong acids are characterized by their ability to completely dissociate into ions when dissolved in water. This complete ionization is why the concentration of the original acid is the same as the concentration of the ions produced.
A common example is nitric acid (HNO\(_3\)), a strong acid used in our problem. It dissociates completely in water:
\[ \text{HNO}_3 + \text{H}_2\text{O} \rightarrow \text{H}_3\text{O}^+ + \text{NO}_3^- \]
In this process:

• Each molecule of HNO\(_3\) results in one hydronium ion, \(\text{H}_3\text{O}^+\).
• This means the molarity of HNO\(_3\) corresponds directly to the molarity of \(\text{H}_3\text{O}^+\) ions.

Understanding the behavior of strong acids helps predict their impact on the solution. Since they fully dissociate, calculations are simplified, unlike with weak acids which only partially ionize, requiring equilibrium calculations to determine concentration.

Chemical Equation Balancing

The first step in understanding any chemical reaction is writing and balancing its chemical equation. This makes it possible to determine the stoichiometry, which is the ratio of reactants and products.In our problem, the balanced chemical equation was:
HNO\(_3\) + H\(_2\)O → \(\text{H}_3\text{O}^+\) + NO\(_3^−\)
Here, one molecule of HNO\(_3\) reacts with a water molecule to form hydronium ions and nitrate ions in a one-to-one ratio. This balancing ensures the "Law of Conservation of Mass" is respected, meaning matter is neither created nor destroyed during the reaction.
When balancing equations:

• Ensure same number of each type of atom on both sides of the equation.
• Adjust coefficients, not subscripts.

Balancing the equation correctly is crucial because it affects concentration calculations directly, giving insight into how many products a given amount of reactant will yield.