Question

Titration of a \(50.00 \mathrm{~mL}\) sample of acid rain required \(9.30 \mathrm{~mL}\) of \(0.0012 \mathrm{M} \mathrm{NaOH}\) to reach the end point. What was the total \(\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]\) in the rain sample? What was the \(\mathrm{pH}\) ?

Short answer

[H₃O⁺] = 2.232 x 10⁻⁴ M; pH ≈ 3.65

Step-by-step solution

Write the balanced chemical equation

The titration process involves the reaction of hydronium ions (\(\text{H}_3\text{O}^+\)) with sodium hydroxide (\(\text{NaOH}\)). The balanced chemical equation is: \[\text{H}_3\text{O}^+ + \text{OH}^- \rightarrow 2 \text{H}_2\text{O}\] Since NaOH provides the hydroxide ions, this equation shows a 1:1 mole ratio between \(\text{H}_3\text{O}^+\) and \(\text{OH}^-\).

Calculate moles of NaOH used

First, calculate the number of moles of \(\text{NaOH}\) used in the titration. The formula for calculating moles is: \[moles = Molarity \times Volume\]Substitute the given values: \[moles = 0.0012 \, \text{mol/L} \times 9.30 \, \text{mL} \times \frac{1 \, \text{L}}{1000 \, \text{mL}} = 0.00001116 \, \text{mol}\]

Determine moles of \(\text{H}_3\text{O}^+\) in the sample

With a 1:1 mole ratio between \(\text{NaOH}\) and \(\text{H}_3\text{O}^+\), the number of moles of \(\text{H}_3\text{O}^+\) is equal to the number of moles of \(\text{NaOH}\): \[moles \text{H}_3\text{O}^+ = 0.00001116 \, \text{mol}\]

Calculate hydronium ion concentration

To find the hydronium ion concentration \([\text{H}_3\text{O}^+]\), divide the moles of \(\text{H}_3\text{O}^+\) by the volume of the acid rain sample in liters:\[[\text{H}_3\text{O}^+] = \frac{0.00001116 \, \text{mol}}{0.05000 \, \text{L}} = 2.232 \times 10^{-4} \, \text{M}\]

Calculate the pH of the solution

The \(\text{pH}\) is calculated using the formula: \[\text{pH} = -\log_{10} [\text{H}_3\text{O}^+]\]Substituting the concentration:\[\text{pH} = -\log_{10} (2.232 \times 10^{-4}) \approx 3.65\]

Key concepts

Hydronium Ion Concentration

In chemistry, hydronium ions (\(\text{H}_3\text{O}^+\)) are significant because they directly influence the acidity of a solution. During a titration, such as with acid rain and sodium hydroxide (\(\text{NaOH}\)), calculating the hydronium ion concentration involves measuring the amount of reactant used to neutralize the acid.

The neutralization reaction occurs according to the equation wherein \(\text{H}_3\text{O}^+\) ions react with hydroxide ions (\(\text{OH}^-\)) from \(\text{NaOH}\) to form water. During the titration, by knowing the amount of \(\text{NaOH}\) added, we determine the moles of \(\text{H}_3\text{O}^+\):

• Calculate the moles of \(\text{NaOH}\), as shown, and assume a 1:1 ratio for \(\text{H}_3\text{O}^+\).
• Divide the moles of \(\text{H}_3\text{O}^+\) by the volume of the sample to find the concentration \([\text{H}_3\text{O}^+]\). In this case, \(\frac{0.00001116 \, \text{mol}}{0.05000 \, \text{L}} = 2.232 \times 10^{-4} \, \text{M}\).

Understanding this concentration helps in identifying the acidity of the rain sample.

pH Calculation

The \(\text{pH}\) is a numeric scale used to specify the acidity or basicity of an aqueous solution. In the case of acid rain titration, once the hydronium ion concentration is known, you can calculate the \(\text{pH}\) of the solution.

The \(\text{pH}\) is simply the negative logarithm (base 10) of the hydronium ion concentration: \[\text{pH} = -\log_{10} [\text{H}_3\text{O}^+]\]

Substituting the calculated \([\text{H}_3\text{O}^+]\) value, we get: \[-\log_{10} (2.232 \times 10^{-4})\approx 3.65\]. The result shows the acidic nature of the rain sample,

• where a \(\text{pH}\) less than 7 indicates acidity, a \(\text{pH}\) greater than 7 indicates basicity, and exactly 7 is neutral.
• This straightforward calculation is pivotal to understanding various environmental effects, such as plant and aquatic life impacted by acid rain.

Thus, determining the \(\text{pH}\) helps elucidate the environmental impact of such solutions.

Balanced Chemical Equation

Balanced chemical equations are fundamental to understanding chemical reactions. In the titration process of acid rain using \(\text{NaOH}\), the reaction is balanced as follows: \[\text{H}_3\text{O}^+ + \text{OH}^- \rightarrow 2 \text{H}_2\text{O}\].

This equation illustrates how every hydronium ion reacts with one hydroxide ion, forming two water molecules.

• The molar relationship in the equation reveals that for every mole of \(\text{OH}^-\) provided by \(\text{NaOH}\), one mole of \(\text{H}_3\text{O}^+\) is neutralized.
• This 1:1 stoichiometric ratio is crucial for the titration analysis, leading to accurate concentration and pH values.

By ensuring all equations are balanced, chemists can maintain consistency and accuracy in experimental results. It allows predictions of how substances will interact, ensuring the titration output to reflect true values.