Problem 100 - Chemistry The Central Science In Si Units

Question

A solution is made by adding \(1.000 \mathrm{~g} \mathrm{Ca}(\mathrm{OH})_{2}(s), 100.0 \mathrm{~mL}\) of \(0.10 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}\), and enough water to make a final volume of \(350.0 \mathrm{~mL}\). Assuming that all of the solid dissolves, what is the \(\mathrm{pH}\) of the final solution?

Step-by-Step Solution

Answer

The pH of the final solution is 12.

Step 1: Write the Reaction Equation

First, identify the chemical reaction occurring in the solution. The reaction between calcium hydroxide and sulfuric acid is: \[ \text{Ca(OH)}_2 + \text{H}_2\text{SO}_4 \rightarrow \text{CaSO}_4 + 2\text{H}_2\text{O} \]This reaction shows that calcium hydroxide will react with sulfuric acid to form calcium sulfate and water.

Step 2: Calculate Moles of Reactants

Next, calculate the number of moles of each reactant present. For \(\text{Ca(OH)}_2\):Molar mass of \(\text{Ca(OH)}_2\) is 74.1 g/mol, and thus:\[ \text{Moles of Ca(OH)}_2 = \frac{1.000 \text{ g}}{74.1 \text{ g/mol}} \approx 0.0135 \text{ moles} \]For \(\text{H}_2\text{SO}_4\):\[ \text{Moles of } \text{H}_2\text{SO}_4 = 0.10 \text{ M} \times 0.100 \text{ L} = 0.0100 \text{ moles} \]

Step 3: Determine the Limiting Reactant

Compare the mole ratio from the balanced equation to determine the limiting reactant. The balanced chemical equation suggests a 1:1 ratio. Since there are more moles of \(\text{Ca(OH)}_2\) than \(\text{H}_2\text{SO}_4\) (0.0135 vs. 0.0100), the \(\text{H}_2\text{SO}_4\) is the limiting reactant.

Step 4: Calculate Excess Moles of \(\text{Ca(OH)}_2\)

Find the moles of \(\text{Ca(OH)}_2\) remaining after the reaction:\[ \text{Excess } \text{Ca(OH)}_2 = 0.0135 - 0.0100 = 0.0035 \text{ moles} \]

Step 5: Calculate Hydroxide Ion Concentration

Since \(\text{Ca(OH)}_2\) provides hydroxide ions, determine the concentration of \(\text{OH}^-\) ions in the solution. The concentration of hydroxide ions from the excess \(\text{Ca(OH)}_2\) can be calculated by:\[ \text{[OH]}^- = \frac{0.0035 \text{ moles}}{0.350 \text{ L}} = 0.010 \text{ M} \]

Step 6: Calculate pOH and pH

Using the concentration of hydroxide ions to determine pOH:\[ \text{pOH} = -\log(0.010) = 2 \]Since \(\text{pH} + \text{pOH} = 14\), solve for pH:\[ \text{pH} = 14 - 2 = 12 \]

Key Concepts

Limiting Reactant

In any chemical reaction, the limiting reactant is the substance that is fully consumed first, ceasing the reaction. It determines how much product can be made. In our original exercise, sulfuric acid (\( \text{H}_2\text{SO}_4 \)) is the limiting reactant.
This is because it has fewer moles compared to calcium hydroxide (\( \text{Ca(OH)}_2 \)).

• We start with 0.0100 moles of \( \text{H}_2\text{SO}_4 \) and 0.0135 moles of \( \text{Ca(OH)}_2 \).
• According to the balanced equation, \( \text{Ca(OH)}_2 \) reacts with \( \text{H}_2\text{SO}_4 \) in a 1:1 ratio.

Since we have fewer moles of \( \text{H}_2\text{SO}_4 \), it limits the amount of product formed.

Reaction Equation

A reaction equation is a representation of a chemical reaction with symbols and formulas. It gives the substances involved, both reactants and products.
The reaction between \( \text{Ca(OH)}_2 \) and \( \text{H}_2\text{SO}_4 \) forms calcium sulfate (\( \text{CaSO}_4 \)) and water (\( \text{H}_2\text{O} \)).

• The equation is: \[ \text{Ca(OH)}_2 + \text{H}_2\text{SO}_4 \rightarrow \text{CaSO}_4 + 2\text{H}_2\text{O} \]
• This shows each molecule of \( \text{Ca(OH)}_2 \) reacts with a molecule of \( \text{H}_2\text{SO}_4 \) to produce a molecule of \( \text{CaSO}_4 \) and two molecules of \( \text{H}_2\text{O} \).

This balanced equation helps identify ratios and atom conservation.

Hydroxide Ion Concentration

The concentration of hydroxide ions (\( \text{OH}^- \)) in a solution can dictate its basicity or alkalinity. Since \( \text{Ca(OH)}_2 \) dissociates to give \( \text{OH}^- \) ions, we calculate its excess to find the \( \text{OH}^- \) concentration.

• From the original exercise, 0.0035 moles of \( \text{Ca(OH)}_2 \) remains unreacted.
• This produces \( 0.0035 \text{ moles} / 0.350 \text{ L} = 0.010 \text{ M} \) \( \text{OH}^- \).

A higher concentration of \( \text{OH}^- \) leads to a higher pH, indicating a more basic solution.

Calcium Hydroxide

Calcium hydroxide, or slaked lime, is a common laboratory reagent. Often, it appears as a white solid or a colorless solution when hydrated. In chemistry, it's known to generate \( \text{OH}^- \) ions, affecting the pH of solutions.

• Its chemical formula is \( \text{Ca(OH)}_2 \) with a molar mass of about 74.1 g/mol.
• When it dissolves, it yields \( \text{Ca}^{2+} \) and \( \text{OH}^- \) ions, making solutions more alkaline.

In the original exercise, it starts as a reactant but some remains after the reaction, impacting the pH.

Sulfuric Acid

Sulfuric acid is a strong and widely used acid in the industry and laboratories. Known for its viscous nature and ability to donate protons, it reacts fully in water.

• The formula is \( \text{H}_2\text{SO}_4 \), and it ionizes to produce \( \text{H}^+ \) ions.
• In our exercise context, it's the limiting reactant.
• This means all \( \text{H}_2\text{SO}_4 \) reacts with \( \text{Ca(OH)}_2 \) before any \( \text{Ca(OH)}_2 \) is left over.

Since \( \text{H}_2\text{SO}_4 \) strongly affects pH, understanding its amount is crucial.