Question

Arsenic acid, \(\mathrm{H}_{3} \mathrm{AsO}_{4}\), is a poisonous acid that has been used in the treatment of wood to prevent insect damage. Arsenic acid has three acidic protons. Say you take a 25.00 -mL sample of arsenic acid and prepare it for titration with \(\mathrm{NaOH}\) by adding \(25.00 \mathrm{~mL}\) of water. The complete neutralization of this solution requires the addition of \(53.07 \mathrm{~mL}\) of \(0.6441 \mathrm{M} \mathrm{NaOH}\) solution. Write the balanced chemical reaction for the titration, and calculate the molarity of the arsenic acid sample.

Short answer

The molarity of arsenic acid is 0.4560 M.

Step-by-step solution

Understand the Reaction

The titration involves arsenic acid, \(\mathrm{H}_3\mathrm{AsO}_4\), and sodium hydroxide, \(\mathrm{NaOH}\). The complete neutralization reaction can be written as:\[\mathrm{H}_3\mathrm{AsO}_4 (aq) + 3\mathrm{NaOH} (aq) \rightarrow \mathrm{Na}_3\mathrm{AsO}_4 (aq) + 3\mathrm{H}_2\mathrm{O} (l)\]This reaction shows that one mole of \(\mathrm{H}_3\mathrm{AsO}_4\) reacts with 3 moles of \(\mathrm{NaOH}\).

Calculate Moles of NaOH Used

To calculate the moles of \(\mathrm{NaOH}\) used, we use the molarity equation:\[\text{Moles of } \mathrm{NaOH} = M \times V = 0.6441 \, \text{M} \times 0.05307 \, \text{L} = 0.03420 \, \text{mol}\]

Determine Moles of Arsenic Acid

From the balanced equation, 3 moles of \(\mathrm{NaOH}\) react with 1 mole of \(\mathrm{H}_3\mathrm{AsO}_4\). Therefore, the moles of \(\mathrm{H}_3\mathrm{AsO}_4\) reacted is:\[\text{Moles of } \mathrm{H}_3\mathrm{AsO}_4 = \frac{0.03420 \, \text{mol}}{3} = 0.01140 \, \text{mol}\]

Calculate Molarity of Arsenic Acid

Molarity is defined as moles of solute per liter of solution. The sample of arsenic acid before dilution was 25.00 mL, or 0.02500 L. Therefore, the molarity \(M\) is calculated as:\[M = \frac{0.01140 \, \text{mol}}{0.02500 \, \text{L}} = 0.4560 \, \text{M}\]

Key concepts

Arsenic Acid

Arsenic acid is a chemical compound with the formula \(\mathrm{H}_{3}\mathrm{AsO}_{4}\). It is classified as a triprotic acid, which means it can donate three protons (\(\mathrm{H}^+\) ions) per molecule in an aqueous solution. This acid has significant applications, including its use as a wood preservative due to its insecticidal properties.

In chemical reactions, arsenic acid can completely dissociate, meaning all three of its acidic protons are neutralized, typically in a series of stepwise reactions. The first dissociation can be represented by \(\mathrm{H}_{3}\mathrm{AsO}_{4} \rightarrow \mathrm{H}_{2}\mathrm{AsO}_{4}^- + \mathrm{H}^+\). Each subsequent dissociation becomes progressively weaker.

As a result, the complete neutralization of arsenic acid in a titration requires a strong base like sodium hydroxide (\(\mathrm{NaOH}\)), which will react with these acidic protons, forming water and salt in the process.

Neutralization Reaction

Neutralization reactions are fundamental reactions in chemistry, wherein an acid and a base react to form water and a salt. In the context of this exercise, arsenic acid (\(\mathrm{H}_3\mathrm{AsO}_4\)) reacts with sodium hydroxide (\(\mathrm{NaOH}\)) to form sodium arsenate (\(\mathrm{Na}_3\mathrm{AsO}_4\)) and water. This can be represented by the balanced equation:

• \(\mathrm{H}_3\mathrm{AsO}_4 (aq) + 3\mathrm{NaOH} (aq) \rightarrow \mathrm{Na}_3\mathrm{AsO}_4 (aq) + 3\mathrm{H}_2\mathrm{O} (l)\)

The coefficients in this balanced equation indicate that three moles of \(\mathrm{NaOH}\) are required to completely neutralize one mole of \(\mathrm{H}_3\mathrm{AsO}_4\). This stoichiometric relationship is crucial when calculating the amounts of reactants needed in such reactions. Understanding these relationships helps in determining the unreacted quantities of reagents in a titration, which is vital for calculating molarity later.

Molarity Calculation

Molarity is a measure of the concentration of a solute in a solution, expressed in moles of solute per liter of solution. It is a crucial concept in titration, as it allows for the determination of the concentration of an unknown solution through a known substance.

In this exercise, the calculation of the molarity of arsenic acid involved several steps:

• First, determine the moles of \(\mathrm{NaOH}\) used, using the formula: \(\text{Moles of } \mathrm{NaOH} = M \times V\), where \(M\) is the molarity and \(V\) is the volume in liters.
• Next, use the stoichiometry from the balanced reaction to find the moles of \(\mathrm{H}_3\mathrm{AsO}_4\). Since three moles of \(\mathrm{NaOH}\) react with one mole of arsenic acid, divide the moles of \(\mathrm{NaOH}\) by 3.
• Finally, calculate the molarity of \(\mathrm{H}_3\mathrm{AsO}_4\) by dividing the moles of arsenic acid by its volume in liters. This will provide the molarity as \(0.4560 \, \text{M}\).

This stepwise approach highlights the importance of understanding the relationships between reactants and products in a neutralization reaction, enabling precise molarity calculations.