Problem 100 - Chemistry: The Central Science in SI Units, Global Edition

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 approximately 12.16.

Step 1: 1. Calculate the moles of Ca(OH)2 and H2SO4

First, find the moles of Ca(OH)2 present in the solution by using its mass and molar mass: Moles of Ca(OH)2 = mass of Ca(OH)2 / molar mass of Ca(OH)2 Moles of Ca(OH)2 = 1.000 g / (40.08 + 2 * (15.999 + 1.008)) g/mol Moles of Ca(OH)2 ≈ 0.0125 mol Next, find the moles of H2SO4 present in the solution by using its concentration and volume: Moles of H2SO4 = concentration of H2SO4 * volume of H2SO4 Moles of H2SO4 = 0.10 M * 100.0 mL * (1 L / 1000 mL) Moles of H2SO4 = 0.010 mol

Step 2: 2. Determine the moles of OH- ions remaining after the reaction

The balanced chemical equation for the reaction between Ca(OH)2 and H2SO4 is: Ca(OH)2(s) + H2SO4(aq) → CaSO4(aq) + 2 H2O(l) From the balanced equation, know that 1 mole of Ca(OH)2 reacts with 1 mole of H2SO4. Since there's an excess of Ca(OH)2, some of it will remain unreacted at the end. Therefore, the final solution will be basic. Calculate the moles of Ca(OH)2 remaining after the reaction: Moles of Ca(OH)2 remaining = moles of Ca(OH)2 - moles of H2SO4 Moles of Ca(OH)2 remaining ≈ 0.0125 - 0.0100 Moles of Ca(OH)2 remaining ≈ 0.0025 mol Now, calculate the total moles of OH- ions in the remaining Ca(OH)2: Moles of OH- ions = 0.0025 mol Ca(OH)2 * 2 moles OH- per 1 mol Ca(OH)2 Moles of OH- ions = 0.0050 mol

Step 3: 3. Calculate the concentration of OH- ions in the final solution

We need the volume of the final solution to find the concentration of OH- ions. Final Volume = 100.0 mL (H2SO4) + 350.0 mL (total volume) - 100.0 mL (initial volume) = 350.0 mL Now, calculate the concentration of OH- ions: Concentration of OH- ions = moles of OH- ions / volume of final solution Concentration of OH- ions = 0.0050 mol / 350.0 mL * (1000 mL / 1 L) Concentration of OH- ions = 0.0143 M

Step 4: 4. Calculate the pOH and pH of the final solution

Use the concentration of OH- ions to find the pOH: pOH = -log10(concentration of OH- ions) pOH = -log10(0.0143) pOH ≈ 1.84 Lastly, use the relationship between pOH and pH to find the pH of the final solution: pH = 14 - pOH pH = 14 - 1.84 pH ≈ 12.16 So the pH of the final solution is approximately 12.16.

Key Concepts

Chemical Reactions

Chemical reactions involve the transformation of substances through the breaking and forming of chemical bonds. This usually results in the conversion of reactants into products, characterized by chemical equations. In our exercise, the reaction involves calcium hydroxide \(\text{Ca(OH)}_2\) and sulfuric acid \(\text{H}_2\text{SO}_4\), resulting in calcium sulfate \(\text{CaSO}_4\) and water \(\text{H}_2\text{O}\).

Moles play a crucial role in balancing chemical equations to ensure the conservation of mass. In this reaction, 1 mole of \(\text{Ca(OH)}_2\) reacts with 1 mole of \(\text{H}_2\text{SO}_4\), indicating a 1:1 stoichiometry. Understanding the balanced equation helps in predicting the amount of reactants consumed and products formed.

To summarize:

• Reactants: \(\text{Ca(OH)}_2\) and \(\text{H}_2\text{SO}_4\)
• Products: \(\text{CaSO}_4\) and \(\text{H}_2\text{O}\)
• Reaction type: Neutralization (acid-base reaction)

Molarity

Molarity is a way to express the concentration of a solution. It's defined as the number of moles of solute per liter of solution, denoted by the unit \( \text{M} \). Molarity is crucial in chemical calculations to determine how concentrated a solution is.

In the given exercise, we had a solution of sulfuric acid \(\text{H}_2\text{SO}_4\) with a molarity of \(0.10 \text{M}\). This means that every liter of this solution contains 0.10 moles of \(\text{H}_2\text{SO}_4\). Using the known molarity and the volume of the solution, we calculate the number of moles of \(\text{H}_2\text{SO}_4\):

\[\text{Moles of } \text{H}_2\text{SO}_4 = \text{Molarity} \times \text{Volume (L)}\]

• This is essential for understanding the reactant quantity available for chemical reactions.
• In earlier steps, this concept allowed us to find that there were 0.010 moles of \(\text{H}_2\text{SO}_4\) available.

Acid-Base Reaction

An acid-base reaction is a specific kind of chemical reaction where an acid reacts with a base, generally producing a salt and water. It's a subtype of neutralization reactions. In our problem, we have \(\text{Ca(OH)}_2\), a base, reacting with \(\text{H}_2\text{SO}_4\), an acid. The balanced chemical equation was essential to determine how much of each reactant is required.

For acid-base reactions, understanding the balanced equation helps in predicting the excess reactant and the one completely consumed. In this instance, since \(\text{Ca(OH)}_2\) is in excess, some remains unreacted, affecting the solution's basicity.

Key aspects of this reaction:

• Acid involved: \(\text{H}_2\text{SO}_4\)
• Base involved: \(\text{Ca(OH)}_2\)
• Product formed: \(\text{CaSO}_4\) and water
• Excess base means a surplus of \(\text{OH}^-\) ions, resulting in a basic solution.

Mole Concept

The mole concept is fundamental in chemistry, allowing chemists to count atoms, molecules, and ions accurately by using the unit 'mole'. A mole contains \(6.022 \times 10^{23}\) entities (Avogadro's number), linking macroscopic measurements to microscopic amounts.

In the exercise you've worked on, calculating moles was a critical step to understand how much \(\text{Ca(OH)}_2\) and \(\text{H}_2\text{SO}_4\) were present and how they reacted. By using mass, volume, and molarity, we were able to calculate that:

• There were initially 0.0125 moles of \(\text{Ca(OH)}_2\) present.
• There were 0.010 moles of \(\text{H}_2\text{SO}_4\).

Understanding this concept enables us to determine the number of \(\text{OH}^-\) ions left unreacted, crucial for calculating the final pH.