Problem 108 - Chemistry: Principles and Reactions

Question

The solubility of \(\mathrm{Ca}(\mathrm{OH})_{2}\) at \(25^{\circ} \mathrm{C}\) is \(0.153 \mathrm{~g} / 100 \mathrm{~g} \mathrm{H}_{2} \mathrm{O}\). Assuming that the density of a saturated solution is \(1.00 \mathrm{~g} / \mathrm{mL}\), calculate the maximum \(\mathrm{pH}\) one can obtain when \(\mathrm{Ca}(\mathrm{OH})_{2}\) is dissolved in water.

Step-by-Step Solution

Answer

The maximum pH of a saturated solution of calcium hydroxide (Ca(OH)2) at 25°C is 12.615.

Step 1: Calculate the molar solubility

First, we need to find the molar solubility of \(\mathrm{Ca}(\mathrm{OH})_{2}\). To do this, we will convert the given solubility in grams into moles per liter. The molar mass of \(\mathrm{Ca}(\mathrm{OH})_{2}\) is \((40.08 + 2 \times (16.00 + 1.008)) \mathrm{g/mol} = 74.096 \mathrm{g/mol}\). Given that the density of the saturated solution is \(1.00 \mathrm{~g/mL}\), the solubility in grams per liter is \(0.153 \mathrm{~g} / 100 \mathrm{~g} \mathrm{H}_{2} \mathrm{O} \times 1000 \mathrm{~g} \mathrm{H}_{2} \mathrm{O} / \mathrm{L} = 1.53 \mathrm{~g/L}\). Now, we can find the molar solubility: Molar solubility \(= \frac{1.53 \mathrm{~g/L}}{74.096 \mathrm{g/mol}} = 0.0206 \mathrm{mol/L}\)

Step 2: Write the balanced equation

Write the balanced equation for the dissolution of \(\mathrm{Ca}(\mathrm{OH})_{2}\): \(\mathrm{Ca}(\mathrm{OH})_{2}(s) \rightleftharpoons \mathrm{Ca}^{2+}(aq) + 2 \mathrm{OH}^-(aq)\)

Step 3: Calculate the concentration of \(\mathrm{OH}^{-}\) ions

Using stoichiometry, we can determine the concentration of \(\mathrm{OH}^-\) ions in the saturated solution. For each mole of \(\mathrm{Ca}(\mathrm{OH})_{2}\) that dissolves, \(2\) moles of \(\mathrm{OH}^-\) ions are produced: Concentration of \(\mathrm{OH}^-\) \(= 2 \times\) (molar solubility of \(\mathrm{Ca}(\mathrm{OH})_{2}\)) \(= 2 \times 0.0206 \mathrm{mol/L} = 0.0412 \mathrm{mol/L}\)

Step 4: Calculate the pOH

Now, we can calculate the pOH of the solution using the formula: pOH \(= -\log_{10}[\mathrm{OH}^-]\) \(= -\log_{10}(0.0412) = 1.385\)

Step 5: Calculate the maximum pH

Finally, we will use the relationship between pH and pOH to find the maximum pH. The relationship is given by: pH + pOH \(= 14\) Substitute the pOH value into the equation and solve for pH: pH \(= 14 - 1.385 = 12.615\) Therefore, the maximum pH that can be obtained when \(\mathrm{Ca}(\mathrm{OH})_{2}\) is dissolved in water is \(12.615\).

Key Concepts

pH Calculation

In this exercise, we aim to find the maximum pH of a solution when calcium hydroxide (\(\text{Ca(OH)}_2\) ) is dissolved in water. Understanding pH is key to determining how acidic or basic a solution is. pH is calculated as\[\text{pH} = 14 - \text{pOH}\]You first need to determine pOH, which represents the concentration of hydroxide ions in the solution. Once you have pOH, you can easily calculate the pHby using the relationship where pH and pOH together equal 14 in a neutral solution at 25°C.

• If the pH is less than 7, the solution is acidic.
• If the pH is exactly 7, the solution is neutral.
• If the pH is greater than 7, the solution is basic.

For calcium hydroxide, which is a strong base, the pH will exceed 7.

Molar Solubility

Molar solubility is critical in this context as it indicates how much solute can dissolve in a solvent to form a saturated solution. For calcium hydroxide, this means finding how many moles of \(\text{Ca(OH)}_2\) can dissolve in a liter of water. Given that the solubility in grams per 100 grams of water is known, we convert this to grams per liter, as the density of the solution is 1 g/mL. This results in a solubility of 1.53 g/L. To find molar solubility, utilize the molar mass of \(\text{Ca(OH)}_2\), which is 74.096 g/mol. Then calculate:\[\text{Molar solubility} = \frac{1.53 \text{ g/L}}{74.096 \text{ g/mol}} = 0.0206 \text{ mol/L}\]This represents the amount of \(\text{Ca(OH)}_2\) that dissolves per liter under the given conditions.

Calcium Hydroxide

Calcium hydroxide is a strong base often used in industrial and chemical applications. It dissociates completely in water, which means it breaks down into its ions:\[\text{Ca(OH)}_2(s) \rightleftharpoons \text{Ca}^{2+}(aq) + 2 \text{OH}^-(aq)\]When dissolved, \(\text{Ca(OH)}_2\) provides one calcium ion and two hydroxide ions per formula unit. The presence of these hydroxide ions is significant because it leads to an increase in the pH of the solution.
Calcium hydroxide's reactivity with acids, forming salts and water, makes it valuable in neutralization processes. Its usage spans construction, wastewater treatment, and food preparation. Understanding its solubility and ionization is vital for applications requiring precise pH control.

Balanced Chemical Equation

A balanced chemical equation is fundamental in chemistry because it represents the conservation of matter. It ensures that the number of atoms for each element is the same on both sides of the reaction. For the dissociation of \(\text{Ca(OH)}_2\), the balanced equation:\[\text{Ca(OH)}_2(s) \rightleftharpoons \text{Ca}^{2+}(aq) + 2 \text{OH}^-(aq)\]indicates that every molecule of dissolved \(\text{Ca(OH)}_2\) yields one calcium ion and two hydroxide ions.
In this particular exercise, one must understand this stoichiometric relationship as it has a direct impact on calculating the pOH and, consequently, the pH. Each mole of \(\text{Ca(OH)}_2\) releasing two moles of \(\text{OH}^-\) is pivotal for determining the basic nature of the solution.