Problem 45 - Chemistry: The Central Science in SI Units, Global Edition

Question

Calculate \(\left[\mathrm{OH}^{-}\right]\) and \(\mathrm{pH}\) for (a) \(1.5 \times 10^{-3} \mathrm{MSr}(\mathrm{OH})_{2},\) (b) \(2.250 \mathrm{~g}\) of \(\mathrm{LiOH}\) in \(250.0 \mathrm{~mL}\) of solution, \((\mathbf{c}) 1.00\) \(\mathrm{mL}\) of \(0.175 \mathrm{M} \mathrm{NaOH}\) diluted to \(2.00 \mathrm{~L},(\mathbf{d})\) a solution formed by adding \(5.00 \mathrm{~mL}\) of \(0.105 \mathrm{M} \mathrm{KOH}\) to \(15.0 \mathrm{~mL}\) of \(9.5 \times 10^{-2} \mathrm{MCa}(\mathrm{OH})_{2}\)

Step-by-Step Solution

Answer

(a) [OH⁻] = 3.0 x 10⁻³ M; pH = 11.48 (b) [OH⁻] = 0.392 M; pH = 13.593 (c) [OH⁻] = 8.75 x 10⁻⁵ M; pH = 9.94 (d) [OH⁻] = 0.16875 M; pH = 13.227

Step 1: (a) Finding [OH⁻] for 1.5 x 10⁻³ M Sr(OH)₂

We are given the molar concentration of Sr(OH)₂, and we know that one mole of Sr(OH)₂ dissociates into two moles of OH⁻ ions. So, the [OH⁻] can be calculated as follows: [OH⁻] = 2 * (1.5 x 10⁻³) = 3.0 x 10⁻³ M

Step 2: (a) Finding pH for 1.5 x 10⁻³ M Sr(OH)₂

First, we need to find the pOH: pOH = -log(3.0 x 10⁻³) = 2.52 Now, we can calculate the pH using the relationship: pH = 14 - pOH = 14 - 2.52 = 11.48

Step 3: (b) Finding [OH⁻] for 2.250 g of LiOH in 250.0 mL of solution

First, we need to calculate the moles of LiOH: Moles of LiOH = mass / molar mass = 2.250 g / (6.94 g/mol + 15.999 g/mol) = 2.250 g / 22.939 g/mol = 0.0981 mol Next, we need to find the molar concentration: [LiOH] = moles / volume = 0.0981 mol / 0.250 L = 0.392 M Since one mole of LiOH dissociates into one mole of OH⁻ ions, the [OH⁻] is equal to the [LiOH]: [OH⁻] = 0.392 M

Step 4: (b) Finding the pH for 2.250 g of LiOH in 250.0 mL of solution

First, we need to find the pOH: pOH = -log(0.392) = 0.407 Now, we can calculate the pH using the relationship: pH = 14 - pOH = 14 - 0.407 = 13.593

Step 5: (c) Finding [OH⁻] for 1.00 mL of 0.175 M NaOH diluted to 2.00 L

First, we need to find the moles of NaOH: Moles of NaOH = volume * concentration = 0.001 L * 0.175 M = 0.000175 mol Next, we need to find the molar concentration after dilution: [OH⁻] = moles / volume = 0.000175 mol / 2.00 L = 8.75 x 10⁻⁵ M

Step 6: (c) Finding the pH for 1.00 mL of 0.175 M NaOH diluted to 2.00 L

First, we need to find the pOH: pOH = -log(8.75 x 10⁻⁵) = 4.06 Now, we can calculate the pH using the relationship: pH = 14 - pOH = 14 - 4.06 = 9.94

Step 7: (d) Finding [OH⁻] for a solution formed by adding 5.00 mL of 0.105 M KOH to 15.0 mL of 9.5 x 10⁻² M Ca(OH)₂

First, we need to find the moles of OH⁻ ions from each compound: Moles OH⁻ from KOH = volume * concentration = 0.005 L * 0.105 M = 5.25 x 10⁻⁴ mol Moles OH⁻ from Ca(OH)₂ = volume * concentration * 2 = 0.015 L * 9.5 x 10⁻² M * 2 = 2.85 x 10⁻³ mol Now, we need to find the total moles of OH⁻ ions and calculate the molar concentration: Total moles OH⁻ = 5.25 x 10⁻⁴ mol + 2.85 x 10⁻³ mol = 3.375 x 10⁻³ mol [OH⁻] = total moles OH⁻ / total volume = 3.375 x 10⁻³ mol / 0.020 L = 0.16875 M

Step 8: (d) Finding the pH for a solution formed by adding 5.00 mL of 0.105 M KOH to 15.0 mL of 9.5 x 10⁻² M Ca(OH)₂

First, we need to find the pOH: pOH = -log(0.16875) = 0.773 Now, we can calculate the pH using the relationship: pH = 14 - pOH = 14 - 0.773 = 13.227

Key Concepts

Hydroxide Ion Concentration

Hydroxide ion concentration is crucial for understanding the basicity of a solution, especially when dealing with strong bases like Sr(OH)2, LiOH, NaOH, and KOH. These compounds dissociate in water to form hydroxide ions, and the concentration of these ions, \( [OH^-] \), dictates the solution's basic strength.

To calculate \( [OH^-] \), you need to understand how each molecule of the base affects the ion concentration:

• For Sr(OH)2, each molecule releases two hydroxide ions, doubling the initial concentration given.
• For bases like LiOH and NaOH, which release one hydroxide ion per molecule, the \( [OH^-] \) is the same as the molar concentration of the base.

Knowing the hydroxide ion concentration allows for the calculation of \( pOH \) and, subsequently, the \( pH \), using the relationships \( pOH = -\log([OH^-]) \) and \( pH = 14 - pOH \). Understanding these concepts ensures you can accurately determine the pH of your solution.

Dilution Calculations

Dilution is a process used to decrease the concentration of a solution, typically by adding more solvent. The principle is based on the conservation of moles, where the total amount of solute (in moles) before and after dilution remains constant.

When performing dilution calculations, you can use the formula: dilution factor \( = \left( \frac{C_1 \times V_1}{C_2 \times V_2} \right) \). Where \( C_1 \) and \( V_1 \) are the initial concentration and volume, and \( C_2 \) and \( V_2 \) are the final concentration and volume.

• For example, calculating \( [OH^-] \) after diluting 1.00 mL of 0.175 M NaOH to 2.00 L involves determining the new concentration using the initial moles of NaOH divided by the final volume.

By understanding the concepts of dilution, you can accurately adjust concentrations for a wide array of chemical experiments.

Molarity Calculations

Molarity is a central concept in solution chemistry that expresses the concentration of a solute in a solution. It is defined as the number of moles of solute per liter of solution, represented as M.

Molarity can be determined by the equation: \( M = \frac{\text{moles of solute}}{\text{volume of solution in liters}} \). Calculating molarity allows for the determination of concentrations needed for further calculations, like those of dilution and hydroxide ion concentration.

• For instance, to find the molarity of a solution prepared from 2.250 g of LiOH in 250 mL, determine the moles of LiOH using its molar mass and divide by the volume in liters.

Understanding and calculating molarity is critical in preparing solutions with precise concentrations for accurate experimental results.

Strong Bases

Strong bases are compounds that completely dissociate into their ions in solution, significantly contributing to the solution's basicity. Reflecting this characteristic, examples of strong bases include Sr(OH)2, LiOH, NaOH, and KOH, which fully ionize to produce hydroxide ions, \( OH^- \).

The strength of these bases is reflected in the high hydroxide ion concentrations they produce, which significantly affects the pH of the solution.

• For example, Sr(OH)2 yields two \( OH^- \) ions per formula unit, so the hydroxide concentration is twice the molarity of the base.
• Similarly, LiOH and NaOH yield one \( OH^- \) per mole, equating the hydroxide concentration to the base's molarity.

Recognizing these characteristics aids in predicting and calculating the \( pH \) of solutions containing strong bases.