91Ó°ÊÓ

When dealing with solutions and their pH levels, knowing how to calculate the pOH is fundamental. The pOH value provides us insights into the concentration of hydroxide ions present in the solution. The relationship between pH and pOH is illustrated by the equation:\[ pH + pOH = 14 \]In our scenario, the solution has a pH of 12.35. To find the pOH, we subtract the pH from 14:\[ pOH = 14 - 12.35 \]\[ pOH = 1.65 \]This simple subtraction tells us the pOH, which plays a crucial role in our next calculations. Understanding this connection between pH and pOH is vital when analyzing acidity and basicity in solutions.

Once you have calculated the pOH of a solution, the next step is to find the hydroxide ion concentration, Denoted as \([OH^-]\), this measurement helps determine how basic the solution is. The ion concentration is found using the formula:\[ [OH^-] = 10^{-pOH} \]With a pOH of 1.65, you can determine the concentration as follows:\[ [OH^-] = 10^{-1.65} \]\[ [OH^-] \approx 0.02238 \text{ M} \]This process transforms a logarithmic pOH value into an actual molar concentration, making it easier to understand the basicity. Understanding hydroxide ion concentration is necessary for further calculations involving solubility and reactions.

The solubility product constant, denoted as \(K_{sp}\), indicates the solubility of a compound under specified conditions. For calcium hydroxide, knowing \(K_{sp}\) helps us comprehend its dissolution behavior.The dissolution of calcium hydroxide can be represented as:\[ Ca(OH)_2 \rightarrow Ca^{2+} + 2OH^- \]For every mole of calcium hydroxide dissolved, one mole of calcium ions and two moles of hydroxide ions are formed. Consequently, the concentration of calcium ions is half that of hydroxide ions:\[ [Ca^{2+}] = \frac{[OH^-]}{2} \]Since the \( [OH^-] \approx 0.02238 \text{ M} \), we find:\[ [Ca^{2+}] \approx 0.01119 \text{ M} \]The \(K_{sp}\) expression for calcium hydroxide is:\[ K_{sp} = [Ca^{2+}][OH^-]^2 \]By substituting the calculated concentrations:\[ K_{sp} = (0.01119)(0.02238)^2 \]\[ K_{sp} \approx 5.61 \times 10^{-6} \]This calculation highlights calcium hydroxide's limited solubility, helping understand its role in various chemical situations.