Question

A solution is prepared by dissolving \(2.00\) moles of \(\mathrm{HNO}_{3}\) in enough water to get \(800.0 \mathrm{~mL}\) of solution. What are the \(\mathrm{H}_{3} \mathrm{O}^{+}\) and the \(\mathrm{OH}\) molar concentrations?

Short answer

The molarity of HNO₃ is \(M = \frac{2.00 \text{ moles}}{0.8 \text{ L}} = 2.50 \text{ M}\). Since HNO₃ is a strong acid, its concentration is equal to the H₃O⁺ concentration: \([H₃O⁺] = 2.50 \text{ M}\). Using the ion product of water, \(K_w = 1.0 \times 10^{-14}\), we find the OH⁻ concentration: \([OH⁻] = \frac{K_w}{[H₃O⁺]} = \frac{1.0 \times 10^{-14}}{2.50} \approx 4.0 \times 10^{-15} \text{ M}\).

Step-by-step solution

Calculate the molarity of HNO₃ solution

To find the molarity, divide the number of moles of solute by the volume of the solution (in liters). The volume of the solution is 800.0 mL, which is 0.8 L. Molarity (M) = moles of solute / volume of solution (L) M = 2.00 moles / 0.8 L

Calculate the H₃O⁺ concentration

Since nitric acid (HNO₃) is a strong acid, it will ionize completely in water: HNO₃ → H⁺ + NO₃⁻ By doing so, it will release H⁺ ions, which will combine with water molecules to form H₃O⁺ ions in the solution. Hence, the concentration of H₃O⁺ ions will be equal to the concentration of HNO₃ in the solution: [H₃O⁺] = M (HNO₃)

Calculate the OH⁻ concentration

Using the relation between H₃O⁺ and OH⁻ concentrations, we can calculate the OH⁻ concentration. This relation is given by the ion product of water (Kw): Kw = [H₃O⁺] × [OH⁻] At 25℃, the value of Kw is \(1.0 \times 10^{-14}\). Rearrange the equation to solve for [OH⁻]: [OH⁻] = Kw / [H₃O⁺] Now we can plug in the values for the molarity of HNO₃, H₃O⁺ concentration, and Kw to calculate the OH⁻ concentration.

Key concepts

Strong Acid

When we talk about strong acids, we're referring to acids that completely dissociate in water. This means every molecule of the acid will split into its ions. For example, nitric acid (\(\text{HNO}_3\)) is a strong acid. When \(\text{HNO}_3\) dissolves in water, it completely ionizes into \(\text{H}^+\) and \(\text{NO}_3^-\). This complete dissociation ensures that the hydrogen ions readily combine with water to form hydronium ions (\(\text{H}_3\text{O}^+\)).

Here are some key points about strong acids:

• They ionize fully in aqueous solutions.
• They produce a high concentration of hydronium ions, which affects the solution's pH level.
• The concentration of hydronium ions in the solution is the same as the concentration of the dissolved strong acid.

Understanding that nitric acid is a strong acid helps us quickly deduce the \(\text{H}_3\text{O}^+\) concentration in our exercise, making it equivalent to the acid's molarity.

Molarity Calculation

Molarity is a measure of the concentration of a solution, expressed as the number of moles of solute per liter of solution. To find the molarity (\(\text{M}\)), you use the formula:\[\text{M} = \frac{\text{moles of solute}}{\text{volume of solution in liters}}\]In the given exercise, you're asked to calculate the molarity of \(\text{HNO}_3\) when 2.00 moles of it is dissolved in 800.0 mL of solution. First, convert the volume from milliliters to liters, since molarity is defined per liter. This conversion is done by dividing milliliters by 1000:

• 800.0 mL = 0.8 L

Now, apply the molarity formula:

• \(\text{M} = \frac{2.00 \text{ moles}}{0.8 \text{ L}}\ = 2.5 \text{ M}\)

This result tells us that the \(\text{HNO}_3\) solution is 2.5 molar, indicating there are 2.5 moles of \(\text{HNO}_3\) in every liter of the solution.

Ion Product of Water

In any aqueous solution, water self-ionizes to a tiny extent into hydrogen ions (\(\text{H}^+\)) and hydroxide ions (\(\text{OH}^-\)). This self-ionization is described by the ion product of water, \(\text{K}_w\). At 25°C, \(\text{K}_w\) has a constant value of \(1.0 \times 10^{-14}\). This means:\[\text{K}_w = [\text{H}_3\text{O}^+] \times [\text{OH}^-] = 1.0 \times 10^{-14}\]Using this relationship, if you know the concentration of \(\text{H}_3\text{O}^+\), you can easily find \(\text{OH}^-\) by rearranging the equation:

• \([\text{OH}^-] = \frac{\text{K}_w}{[\text{H}_3\text{O}^+]}\)

For example, using our previously found \(\text{H}_3\text{O}^+\) concentration of 2.5 M:

• \([\text{OH}^-] = \frac{1.0 \times 10^{-14}}{2.5} \approx 4.0 \times 10^{-15}\)
  

This value shows how the presence of a strong acid like \(\text{HNO}_3\) shifts the balance, reducing the concentration of hydroxide ions significantly, thereby affecting the acidity of the solution.