Problem 143 - General Chemistry

Question

Calculate the \(\mathrm{pH}\) of a solution made by mixing 7.52 \(\mathrm{mL}\) of \(4.9 \times 10^{-2} \mathrm{M} \mathrm{Ca}(\mathrm{OH})_{2}\) with \(22.5 \mathrm{~mL}\) of \(0.11 \mathrm{M} \mathrm{HCl}\)

Step-by-Step Solution

Answer

The pH of the solution is approximately 1.24.

Step 1: Calculate Moles of Each Reactant

First, calculate the number of moles of \( ext{Ca(OH)}_2\) and \( ext{HCl}\). The formula is \( ext{moles} = ext{concentration} \times ext{volume}\). For \( ext{Ca(OH)}_2\):\[moles = 4.9 \times 10^{-2} \, ext{M} \times 7.52 \, ext{mL} \times \frac{1 \, ext{L}}{1000 \, ext{mL}} = 3.69 \times 10^{-4} \, ext{mol}\]For \( ext{HCl}\):\[moles = 0.11 \, ext{M} \times 22.5 \, ext{mL} \times \frac{1 \, ext{L}}{1000 \, ext{mL}} = 2.475 \times 10^{-3} \, ext{mol}\]

Step 2: Determine Limiting Reactant

Determine which reactant is limiting. The reaction is:\[ ext{Ca(OH)}_2 + 2 ext{HCl} \rightarrow 2 ext{H}_2 ext{O} + ext{CaCl}_2\]Since 1 mole of \( ext{Ca(OH)}_2\) reacts with 2 moles of \( ext{HCl}\), the required moles of \( ext{HCl}\) are:\[3.69 \times 10^{-4} \, ext{mol} \times 2 = 7.38 \times 10^{-4} \, ext{mol}\]Since only \(2.475 \times 10^{-3} \, ext{mol}\) \( ext{HCl}\) is available, \( ext{Ca(OH)}_2\) is the limiting reactant.

Step 3: Calculate Remaining HCl after Reaction

After \( ext{Ca(OH)}_2\) completely reacts, calculate the remaining \( ext{HCl}\) by subtracting the moles required:\[2.475 \times 10^{-3} \, ext{mol} - 7.38 \times 10^{-4} \, ext{mol} = 1.737 \times 10^{-3} \, ext{mol}\]

Step 4: Calculate Concentration of Remaining HCl

Calculate the concentration of the remaining \( ext{HCl}\) in the new solution volume (sum of both solution volumes):\[ ext{Total volume} = 7.52 \, ext{mL} + 22.5 \, ext{mL} = 30.02 \, ext{mL} = 0.03002 \, ext{L}\]Concentration of \( ext{HCl}\):\[ ext{Concentration} = \frac{1.737 \times 10^{-3} \, ext{mol}}{0.03002 \, ext{L}} = 0.0579 \, ext{M}\]

Step 5: Calculate pH of the Solution

To find the \( ext{pH}\), use the formula \( ext{pH} = - ext{log}[ ext{H}^+]\), where \([ ext{H}^+]\) is the concentration of \( ext{HCl}\):\[ ext{pH} = - ext{log}(0.0579) \approx 1.24\]

Key Concepts

Limiting Reactant

In any chemical reaction, it's important to know which reactant will run out first. This reactant is known as the "limiting reactant." The limiting reactant determines the amount of product that can be formed.
When we mix calcium hydroxide, \( ext{Ca(OH)}_2\), and hydrochloric acid, \( ext{HCl}\), it's a simple acid-base reaction. By calculating the moles of each reactant using the formula \(\text{{moles}} = \text{{concentration}} \times \text{{volume}}\), we can identify the limiting reactant.

• For \( ext{Ca(OH)}_2\), it’s calculated as \(4.9 \times 10^{-2} \, \text{M} \times 7.52 \, \text{mL} = 3.69 \times 10^{-4} \, \text{mol}\).
• For \( ext{HCl}\), it’s \(0.11 \, \text{M} \times 22.5 \, \text{mL} = 2.475 \times 10^{-3} \, \text{mol}\).

By comparing stoichiometrically, 1 mole of \( ext{Ca(OH)}_2\) reacts with 2 moles of \( ext{HCl}\). We find that \( ext{Ca(OH)}_2\) becomes the limiting reactant as it runs out first, consuming less than 2 moles of \( ext{HCl}\) available for each mole of \( ext{Ca(OH)}_2\) present.

Chemical Reactions

Chemical reactions involve the transformation of reactants into products. In the given example, \( ext{Ca(OH)}_2\) and \( ext{HCl}\) undergo a chemical reaction to form water and calcium chloride \(\text{CaCl}_2\).
It's essential to understand how reactants contribute to product formation. The balanced chemical equation for the reaction provides this relationship:
\[\text{Ca(OH)}_2 + 2\text{HCl} \rightarrow 2\text{H}_2\text{O} + \text{CaCl}_2\]
This balanced equation tells us that each mole of \(\text{Ca(OH)}_2\) reacts with 2 moles of \(\text{HCl}\) to produce 1 mole of calcium chloride along with 2 moles of water.
Understanding this balanced equation is vital to calculate and predict the outcomes in chemical reactions accurately.

Acid-Base Reaction

An acid-base reaction is a type of chemical reaction that involves the neutralization of an acid and a base. In this scenario, \( ext{HCl}\) acts as the acid and \(\text{Ca(OH)}_2\) acts as the base. Acid-base reactions are vital because they're commonly used in pH calculations, making it important to understand their mechanism.
When \( ext{HCl}\) is added to \(\text{Ca(OH)}_2\), they neutralize each other. This neutralization results in the production of water and a salt, which is \(\text{CaCl}_2\) in this case. The reaction can be considered as:

• \