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Understanding how to determine the pH of a solution is crucial in many areas of chemistry and biology. In this exercise, the focus is on understanding the relation between pH and pOH. - **pH** is a measure of how acidic or basic a solution is. It is based on the concentration of hydrogen ions (\(\text{H}^+\)). A lower pH means more acidic, and a higher pH means more basic.- The formula, \(\text{pH} + \text{pOH} = 14\), relates pH and pOH, which helps us find the concentration of hydroxide ions (\(\text{OH}^-\).Given a pH value, as in this problem, we can find the pOH by re-arranging this formula:1. **Substitute the pH into the formula**: For a solution with a pH of 10.05, we have: - 10.05 + pOH = 14 2. **Solve for pOH**: Subtract the pH from 14 to find the pOH: - pOH = 14 - 10.05 = 3.95 This calculation is the first step in determining the concentration of hydroxide ions in the solution.

Calculating the concentration of hydroxide ions (\([\text{OH}^-]\)) is essential for understanding the basicity of a solution. Once you have the pOH from the previous step:- The pOH is related to the concentration of OH鈦� ions through a logarithmic relation: - \(\text{pOH} = -\log_{10}{[\text{OH}^-]}\)To find the concentration, you need to reverse this relationship by using antilogarithms:1. **Use the antilog equation**: Rearrange to solve for \([\text{OH}^-]\): - \([\text{OH}^-] = 10^{-\text{pOH}}\)2. **Substitute the pOH value**: For pOH of 3.95: - \([\text{OH}^-] = 10^{-3.95} \approx 1.12 \times 10^{-4} \text{ M}\)This represents the molarity of hydroxide ions in your solution, a crucial step in determining how basic the solution is.

Stoichiometry helps you understand the relationships between reactants and products in a chemical reaction. For the case of calcium hydroxide, \(\text{Ca(OH)}_2\), each unit dissociates into one calcium ion (\(\text{Ca}^{2+}\)) and two hydroxide ions (\(\text{2OH}^-\)).- **Dissociation reaction**: - \(\text{Ca(OH)}_2 \rightarrow \text{Ca}^{2+} + 2\text{OH}^-\)Given the concentration of \([\text{OH}^-] = 1.12 \times 10^{-4} \text{ M}\), you can find the concentration of \(\text{Ca(OH)}_2\):1. **Use stoichiometry**: For each molecule of \(\text{Ca(OH)}_2\), two hydroxide ions are produced. - Therefore, the concentration of \(\text{Ca(OH)}_2\) is half the concentration of \([\text{OH}^-]\).2. **Calculate concentration**: - \([\text{Ca(OH)}_2] = \frac{1.12 \times 10^{-4} \text{ M}}{2} \approx 5.6 \times 10^{-5} \text{ M}\)Knowing the concentration of \(\text{Ca(OH)}_2\) allows you to further explore the properties of the solution, such as assessing its capacity to neutralize acids.