Problem 103 - Chemistry The Central Science In Si Units

Question

(a) A caesium hydroxide solution is prepared by dissolving \(3.20 \mathrm{~g}\) of \(\mathrm{CsOH}\) in water to make \(25.00 \mathrm{~mL}\) of solution. What is the molarity of this solution? (b) Then, the caesium hydroxide solution prepared in part (a) is used to titrate a hydroiodic acid solution of unknown concentration. Write a balanced chemical equation to represent the reaction between the caesium hydroxide and hydroiodic acid solutions. (c) If \(18.65 \mathrm{~mL}\) of the caesium hydroxide solution was needed to neutralize a \(42.3 \mathrm{~mL}\) aliquot of the hydroiodic acid solution, what is the concentration (molarity) of the acid?

Step-by-Step Solution

Answer

(a) CsOH solution molarity is 0.8536 M. (b) Balanced equation: CsOH + HI → CsI + H2O. (c) HI molarity is 0.376 M.

Step 1: Calculate Moles of CsOH

To find the molarity, we first need the moles of CsOH. The molar mass of CsOH is approximately 149.91 g/mol. Use the formula:\[\text{Moles of } \text{CsOH} = \frac{\text{Mass}}{\text{Molar Mass}} = \frac{3.20 \text{ g}}{149.91 \text{ g/mol}} \approx 0.02134 \text{ mol.}\]

Step 2: Determine the Molarity of CsOH Solution

Molarity is defined as moles of solute per liter of solution. Convert 25.00 mL to liters:\[25.00 \text{ mL} = 0.02500 \text{ L}.\]Then, calculate the molarity:\[\text{Molarity} = \frac{\text{Moles of CsOH}}{\text{Volume of solution in L}} = \frac{0.02134 \text{ mol}}{0.02500 \text{ L}} = 0.8536 \text{ M}.\]

Step 3: Write the Balanced Chemical Equation

The reaction between caesium hydroxide (CsOH) and hydroiodic acid (HI) is a neutralization reaction:\[\text{CsOH} + \text{HI} \rightarrow \text{CsI} + \text{H}_2\text{O}.\]

Step 4: Use Titration Data to Find Moles

We can use the stoichiometry of the reaction (1:1 ratio) to find moles of HI based on the moles of CsOH used. First, find moles of CsOH used in the titration:\[\text{Moles of CsOH} = \text{Molarity} \times \text{Volume in L}\]\[ = 0.8536 \text{ M} \times 0.01865 \text{ L} = 0.01591 \text{ mol.}\]Thus, moles of HI used is also 0.01591 mol.

Step 5: Calculate Molarity of HI

Finally, calculate the molarity of HI. Convert the 42.3 mL of HI to liters:\[42.3 \text{ mL} = 0.0423 \text{ L}\]Now, calculate the molarity:\[\text{Molarity of HI} = \frac{\text{Moles of HI}}{\text{Volume in L}} = \frac{0.01591 \text{ mol}}{0.0423 \text{ L}} = 0.376\text{ M}.\]

Key Concepts

Titration

Titration is an analytical technique used to determine the concentration of a solution by reacting it with a solution of known concentration. This is typically done by using a reagent, known as the titrant, which has a precise molarity, to react with a substance in a solution. In our exercise, caesium hydroxide solution, with a known molarity, acts as the titrant. It is added to a hydroiodic acid solution of unknown concentration until the reaction reaches its endpoint.

During a titration, the volume of the titrant is measured precisely, allowing us to determine the amount of solute in the unknown solution. This is pivotal in calculating the concentration of the unknown acid by utilizing the relationship established through stoichiometry. Titrations are often visualized through a color change in the solution with an indicator or by reaching a specific pH if a pH meter is used.

Balanced Chemical Equation

A balanced chemical equation is crucial in understanding chemical reactions. It depicts the reactants transforming into products with the same number of each type of atom on both sides of the equation. For the reaction in our exercise, caesium hydroxide and hydroiodic acid are the reactants, while cesium iodide and water are the products.

A balanced chemical equation reflects the conservation of mass and charge, showing that atoms are neither created nor destroyed in a chemical reaction. Specifically, the balanced equation for this reaction is:

• CsOH + HI → CsI + H₂O

Each side of the reaction has one Cs, one I, one O, and two H atoms, satisfying the conservation laws. Understanding how to balance such chemical equations is essential for calculating reaction stoichiometry accurately.

Neutralization Reaction

A neutralization reaction is a type of chemical reaction where an acid and a base react to form water and a salt. This effectively neutralizes the acidic and basic properties of the reactants. In our case, caesium hydroxide (CsOH), a base, neutralizes hydroiodic acid (HI), an acid.

The general form of a neutralization reaction can be represented as:

• Acid + Base → Salt + Water

This type of reaction typically produces water, which is the case for the reaction between CsOH and HI, forming water and cesium iodide (CsI) as the salt. Understanding these reactions is essential in titration, as it enables you to predict the products formed and calculate the concentration of unknown solutions.

Stoichiometry

Stoichiometry involves using balanced chemical equations to calculate the relative quantities of reactants and products in a chemical reaction. This concept is based on the conservation of mass and the mole ratio derived from the balanced equation.

In our exercise, stoichiometry comes into play by using the mole ratio from the balanced chemical equation to determine the moles of hydroiodic acid that reacted with a known amount of caesium hydroxide during titration. Since the reaction is in a 1:1 mole ratio (one mole of CsOH reacts with one mole of HI), the moles of CsOH used directly tells us the moles of HI in the sample.

Stoichiometry allows for the precise calculation of concentrations and can be applied to quantify substances, analyze reaction yields, and predict the amounts necessary for reactions, making it an indispensable tool in chemistry.