91Ó°ÊÓ

Calculating the pH of a solution is crucial in understanding its acidity or basicity. pH is related to pOH, where their sum is always 14 at 25 degrees Celsius. To find the pH from the pOH, you use the formula:
\[ \mathrm{pOH} = -\log([ ext{OH}^-]) \]
Given \([\text{OH}^-] = 3.4 \times 10^{-4}\), simply input this into the pOH formula to find:
\(\mathrm{pOH} = -\log(3.4 \times 10^{-4}) \approx 3.47\).
Finally, shorten the journey to pH by applying the well-known relationship:

• \(\mathrm{pH} = 14 - \mathrm{pOH}\)

Given \(\mathrm{pOH} = 3.47\), the solution's pH comes out to:\(\mathrm{pH} = 14 - 3.47 = 10.53\).
Remember, a pH greater than 7 indicates a basic solution, which fits since you are dealing with magnesium hydroxide.

The solubility expression helps in understanding how a compound dissolves in water. For a salt like \(\mathrm{Mg(OH)}_2\), we express its solubility product, \(K_{sp}\), which is a constant at a given temperature. When \(\mathrm{Mg(OH)}_2\) dissolves, it breaks down into ions:

• One \(\mathrm{Mg}^{2+}\)
• Two \(\mathrm{OH}^-\)

This leads us to the formation of the solubility expression:
\[ K_{sp} = [\text{Mg}^{2+}] \cdot [\text{OH}^-]^2 \]
For magnesium hydroxide, \([\text{Mg}^{2+}] = s\) and \([\text{OH}^-] = 2s\) because of the 1:2 ratio. Place these values into the expression and you're effectively on your way to solving the entire problem. It's the framework for understanding how compounds behave in solution.

The calculation of \(K_{sp}\) is vital when predicting if a precipitate will form in a specific solution. For \(\mathrm{Mg(OH)}_2\), the value is given as:
\(K_{sp} = 1.96 \times 10^{-11}\).
This means for saturated solutions at this temperature, the product of the concentrations of the resultant ions is a constant. The dissociation of \(\mathrm{Mg(OH)}_2\) gives the expression:
\(s \cdot (2s)^2 = 1.96 \times 10^{-11}\).

• Simplified to: \(4s^3 = 1.96 \times 10^{-11}\)

To find \(s\), solve:
\(s^3 = \frac{1.96 \times 10^{-11}}{4} = 4.9 \times 10^{-12}\).
The cube root gives \(s = \sqrt[3]{4.9 \times 10^{-12}} \approx 1.7 \times 10^{-4}\), describing the molar solubility.

Understanding the concentration of hydroxide ions in a solution provides insight into its basic properties. For magnesium hydroxide's dissociation:

• \(\text{Mg(OH)}_2 \rightarrow \text{Mg}^{2+} + 2\text{OH}^-\)

The molar ratio is key: each molecule of \(\mathrm{Mg(OH)}_2\) produces two hydroxide ions. If the molarity of \(\mathrm{Mg}^{2+}\) is \(s\), then \([\text{OH}^-] = 2s\).
After solving for \(s\) gives us:
\([\text{OH}^-] = 2 \times 1.7 \times 10^{-4} = 3.4 \times 10^{-4} \text{ M}\).
Grasping [\text{OH}^-] is a cornerstone in assessing solution's character, feeding directly into calculations for other properties like pH.