Problem 46 - Chemistry The Central Science

Question

Calculate \(\left[\mathrm{OH}^{-}\right]\) and \(\mathrm{pH}\) for each of the following strong base solutions: (a) \(0.182 \mathrm{M} \mathrm{KOH},\) (b) \(3.165 \mathrm{~g}\) of \(\mathrm{KOH}\) in \(500.0 \mathrm{~mL}\) of solution, (c) \(10.0 \mathrm{~mL}\) of \(0.0105 \mathrm{M} \mathrm{Ca}(\mathrm{OH})_{2}\) diluted to \(500.0 \mathrm{~mL}\) (d) a solution formed by mixing \(20.0 \mathrm{~mL}\) of \(0.015 \mathrm{M} \mathrm{Ba}(\mathrm{OH})_{2}\) with \(40.0 \mathrm{~mL}\) of \(8.2 \times 10^{-3} \mathrm{M} \mathrm{NaOH}\).

Step-by-Step Solution

Answer

(a) [OH⁻] = 0.182 M; pH = 13.26 (b) [OH⁻] = 0.1128 M; pH = 13.053 (c) [OH⁻] = 4.2 × 10⁻⁴ M; pH = 10.62 (d) [OH⁻] = 1.5467 × 10⁻² M; pH = 12.19

Step 1: Calculate the moles of OH⁻ ions

KOH is a strong base that dissociates completely in water according to the equation: KOH → K⁺ + OH⁻. Therefore, 1 mole of KOH produces 1 mole of OH⁻. The concentration of OH⁻ ions is the same as that of KOH: 0.182 M.

Step 2: Calculate pH

Now we'll use the OH⁻ concentration to calculate the pH using the formula: pOH = -log[OH⁻] and pH = 14 - pOH. pOH = -log(0.182) = 0.74 pH = 14 - 0.74 = 13.26 (a) [OH⁻] = 0.182 M; pH = 13.26 (b) 3.165 g of KOH in 500.0 mL of solution

Step 3: Calculate the moles of OH⁻ ions

First, we'll calculate the moles of KOH by dividing the mass by the molar mass of KOH (56.11 g/mol). moles of KOH = 3.165 g / 56.11 g/mol = 0.0564 mol Since 1 mole KOH produces 1 mole OH⁻, moles of OH⁻ = moles of KOH = 0.0564 mol

Step 4: Calculate the concentration of OH⁻ ions

To find the concentration of OH⁻ ions, divide the moles of OH⁻ by the volume of the solution in liters. [OH⁻] = 0.0564 mol / 0.5 L = 0.1128 M

Step 5: Calculate pH

pOH = -log(0.1128) = 0.947 pH = 14 - 0.947 = 13.053 (b) [OH⁻] = 0.1128 M; pH = 13.053 (c) 10.0 mL of 0.0105 M Ca(OH)₂ diluted to 500.0 mL

Step 6: Calculate the moles of OH⁻ ions

The dissociation of Ca(OH)₂ in water: Ca(OH)₂ → Ca²⁺ + 2OH⁻. Thus, 1 mole of Ca(OH)₂ produces 2 moles of OH⁻ ions. moles of Ca(OH)₂ = 0.0105 M × 0.01 L = 1.05 × 10⁻⁴ moles of OH⁻ = 1.05 × 10⁻⁴ × 2 = 2.1 × 10⁻⁴

Step 7: Calculate the concentration of OH⁻ ions

[OH⁻] = 2.1 × 10⁻⁴ mol / 0.5 L = 4.2 × 10⁻⁴ M

Step 8: Calculate pH

pOH = -log(4.2 × 10⁻⁴) = 3.38 pH = 14 - 3.38 = 10.62 (c) [OH⁻] = 4.2 × 10⁻⁴ M; pH = 10.62 (d) A solution formed by mixing 20.0 mL of 0.015 M Ba(OH)₂ with 40.0 mL of 8.2 × 10⁻³ M NaOH.

Step 9: Calculate the moles of OH⁻ ions

The dissociation of Ba(OH)₂: Ba(OH)₂ → Ba²⁺ + 2OH⁻. Thus, 1 mole of Ba(OH)₂ produces 2 moles of OH⁻ ions. moles of OH⁻ from Ba(OH)₂ = 0.015 M × 0.02 L × 2 = 6 × 10⁻⁴ mol moles of OH⁻ from NaOH = 8.2 × 10⁻³ M × 0.04 L = 3.28 × 10⁻⁴ Total moles of OH⁻ = 6 × 10⁻⁴ + 3.28 × 10⁻⁴ = 9.28 × 10⁻⁴ mol

Step 10: Calculate the concentration of OH⁻ ions

Total volume = 20 mL + 40 mL = 60 mL = 0.06 L [OH⁻] = 9.28 × 10⁻⁴ mol / 0.06 L = 1.5467 × 10⁻² M

Step 11: Calculate pH

pOH = -log(1.5467 × 10⁻²) = 1.81 pH = 14 - 1.81 = 12.19 (d) [OH⁻] = 1.5467 × 10⁻² M; pH = 12.19

Key Concepts

KOH

Potassium hydroxide (KOH) is a common example of a strong base. When KOH dissolves in water, it fully dissociates into potassium ions (K⁺) and hydroxide ions (OH⁻). This complete dissociation is a characteristic feature of strong bases.

• You can easily calculate the concentration of OH⁻ ions; it is the same as the initial concentration of KOH.
• For example, if you have a solution with 0.182 M KOH, the concentration of OH⁻ ions is also 0.182 M.

Consequently, calculating pH also becomes straightforward because it requires converting OH⁻ concentration to pOH and then to pH.

Ca(OH)₂

Calcium hydroxide ( Ca(OH)₂ ) is another potent base, frequently utilized in various applications such as construction. Unlike KOH, Ca(OH)₂ dissociates in a two-step process:

• Each molecule of Ca(OH)₂ produces two hydroxide ions ( OH⁻ ) when it dissolves in water.

For instance, in a solution of 0.0105 M Ca(OH)₂, each mole of Ca(OH)₂ releases two moles of OH⁻ ions, leading to 0.021 M OH⁻ . Understanding this multiplication factor is essential when calculating the net OH⁻ concentration for pH calculations.

Ba(OH)₂

Barium hydroxide, Ba(OH)₂, shares similarities with Ca(OH)₂ in that it also dissociates to form more than one hydroxide ion. Specifically, Ba(OH)₂ dissociates as follows:

• One mole of Ba(OH)₂ produces two moles of OH-).

This is a key point for calculating pH, because it significantly affects the concentration of hydroxide ions in the solution.
Considering the total dissociation in the solution, you multiply the concentration of Ba(OH)₂ by two to find the total OH⁻ concentration.

Strong Base Dissociation

Strong bases, like KOH, Ca(OH)₂, and Ba(OH)₂, are unique in their full dissociation into ions in aqueous solutions. This complete breakdown underlies their ability to significantly increase OH⁻ concentration if they are present.

• For KOH, this means that each molecule turns into one hydroxide ion.
• For Ca(OH)₂ and Ba(OH)₂, each molecule results in two hydroxide ions.

Understanding this full dissociation principle is pivotal because it directly influences the pH, positioning these solutions as among the most basic when dissolved in water.

pH and pOH Relationship

The relationship between pH and pOH is a fundamental concept for understanding acid-base chemistry. Both values measure different aspects of hydrogen and hydroxide ion concentrations in solutions:

• pOH is calculated using the negative logarithm of the hydroxide ion concentration, [OH⁻].
• Then, calculate pH by subtracting pOH from 14 (i.e., pH = 14 - pOH).

This relationship is vital in assessing the acidity or basicity of a solution. The higher the concentration of hydroxide ions, the lower the pOH, and consequently, the higher the pH value, indicating a more basic (or alkaline) solution. This interplay provides insight into the chemical nature of strong bases.