Problem 130 - General Chemistry

Question

Arsenic acid, \(\mathrm{H}_{3} \mathrm{~A} \mathrm{~s} \mathrm{O}_{4}\), 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 26.14-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.

Step-by-Step Solution

Answer

The balanced equation is: \( \text{H}_3\text{AsO}_4 + 3\text{NaOH} \rightarrow \text{Na}_3\text{AsO}_4 + 3\text{H}_2\text{O} \). The molarity of arsenic acid is approximately 0.436 M.

Step 1: Write the Balanced Chemical Reaction

The balanced chemical reaction for the complete neutralization of arsenic acid with sodium hydroxide is as follows:\[ \text{H}_3\text{AsO}_4 (aq) + 3\text{NaOH} (aq) \rightarrow \text{Na}_3\text{AsO}_4 (aq) + 3\text{H}_2\text{O} (l) \]This equation indicates that one mole of arsenic acid reacts with three moles of sodium hydroxide to produce sodium arsenate and water.

Step 2: Calculate Moles of NaOH Used

To find the number of moles of \(\text{NaOH}\), use its volume and concentration:\[ \text{Moles of NaOH} = M(NaOH) \times V(NaOH) = 0.6441 \text{ M} \times 0.05307 \text{ L} = 0.0342 \text{ moles NaOH} \]

Step 3: Determine Moles of Arsenic Acid

According to the balanced chemical equation, every mole of \(\text{H}_3\text{AsO}_4\) reacts with 3 moles of \(\text{NaOH}\). Therefore, the number of moles of \(\text{H}_3\text{AsO}_4\) is:\[ \text{Moles of } \text{H}_3\text{AsO}_4 = \frac{\text{Moles of NaOH}}{3} = \frac{0.0342}{3} = 0.0114 \text{ moles } \text{H}_3\text{AsO}_4 \]

Step 4: Calculate the Molarity of Arsenic Acid

The molarity of the arsenic acid can be calculated using the moles of arsenic acid and the volume of the arsenic acid solution prior to dilution:\[ \text{Volume of } \text{H}_3\text{AsO}_4 = 0.02614 \text{ L} \]\[ \text{Molarity of } \text{H}_3\text{AsO}_4 = \frac{\text{Moles of } \text{H}_3\text{AsO}_4}{\text{Volume of } \text{H}_3\text{AsO}_4} = \frac{0.0114}{0.02614} \approx 0.436 \text{ M} \]

Key Concepts

Balanced Chemical Equation

When dealing with acid-base titrations, writing the balanced chemical equation is the first and crucial step. A balanced equation ensures that the chemical reaction accurately represents the conversion of reactants to products.
In our case, arsenic acid ( \(\text{H}_3\text{AsO}_4\)) reacts with sodium hydroxide ( \(\text{NaOH}\)). This is a typical neutralization reaction where an acid and a base interact to form a salt and water. The balanced equation is:

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

This tells us that one mole of arsenic acid will react with three moles of sodium hydroxide, producing one mole of sodium arsenate and three moles of water. Writing the balanced chemical equation helps in calculating the stoichiometric relationships, which are crucial for further steps like finding the molarity.

Molarity Calculation

Molarity is a measure of concentration, expressed in moles of solute per liter of solution. To calculate the molarity of arsenic acid, we must know the number of moles of arsenic acid and the volume of the solution before dilution.
First, we determine the number of moles of sodium hydroxide used in the titration, since it directly helps us calculate moles of arsenic acid due to their stoichiometric relationship. For sodium hydroxide:

• Molarity = \(0.6441\, \text{M}\)
• Volume = \(53.07\, \text{mL} = 0.05307\, \text{L}\)
• Moles of \(\text{NaOH} = 0.6441 \, \text{M} \times 0.05307\, \text{L} = 0.0342\, \text{moles}\)

Since three moles of \(\text{NaOH}\) are needed for one mole of \(\text{H}_3\text{AsO}_4\), the moles of arsenic acid are calculated as:

• Moles of \(\text{H}_3\text{AsO}_4 = \frac{0.0342}{3} = 0.0114\, \text{moles}\)

Now, using the volume of arsenic acid sample before adding water:

• Volume = \(26.14\, \text{mL} = 0.02614\, \text{L}\)

The molarity (\(M\)) of the arsenic acid is then:

• \(M = \frac{0.0114}{0.02614} \approx 0.436\, \text{M}\)

This provides the concentration of arsenic acid in the solution.

Neutralization Reaction

Neutralization reactions are a type of chemical reaction between an acid and a base that results in the formation of water and a salt. This is a key concept in titration processes.
The main goal of a neutralization reaction is to equilibrate the hydrogen ions from the acid with the hydroxide ions from the base, thus forming water. In this specific exercise, arsenic acid, an acid with three acidic protons, interacts completely with sodium hydroxide:

• \(\text{H}_3\text{AsO}_4\) provides hydrogen ions \(\text{H}^+\)
• \(\text{NaOH}\) supplies hydroxide ions \(\text{OH}^-\)

Each \(\text{H}^+\) from the acid pairs with an \(\text{OH}^-\) from the base, forming water. This balance is captured by the equation:

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

The neutralization reaction is complete when all the acidic protons have reacted with the base, which is indicated by the reaction requiring 53.07 mL of sodium hydroxide here. Understanding this helps in determining the completion of the titration as well as the necessary components involved.