Question

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

Short answer

The short answers for each of the strong base solutions are: a) [OH⁻] = 0.182 M and pH ≈ -0.74 b) [OH⁻] = 0.113 M and pH ≈ -0.95 c) [OH⁻] ≈ 4.2 × 10⁻⁴ M and pH ≈ -3.38 d) [OH⁻] ≈ 1.55 × 10⁻² M and pH ≈ -1.81

Step-by-step solution

Find [OH⁻] concentration

Since KOH is a strong base, we know that [OH⁻] = [KOH]. Therefore, [OH⁻] = 0.182 M.

Calculate the pH

We can convert the [OH⁻] to pH using the formula: pH = -log[OH⁻] = -log(0.182) ≈ -0.74. #b.# Calculate [OH⁻] and pH for 3.165 g of KOH in 500.0 mL of solution

Calculate the molar concentration

First, we need to convert the mass of KOH into moles. moles of KOH = (3.165 g) / (56.11 g/mol) ≈ 0.0564 mol. Now, we can calculate the molar concentration: M = moles / volume = 0.0564 mol / 0.5 L ≈ 0.113 M.

Find [OH⁻] concentration

Since KOH is a strong base, [OH⁻] = [KOH]. Therefore, [OH⁻] = 0.113 M.

Calculate the pH

We can convert the [OH⁻] to pH using the formula: pH = -log[OH⁻] = -log(0.113) ≈ -0.95. #c.# Calculate [OH⁻] and pH for 10.0 mL of 0.0105 MCa(OH)₂ diluted to 500.0 mL

Calculate the final concentration

Since the 10 mL of 0.0105 M solution was diluted, we need to use the dilution formula: M1V1 = M2V2. Solving for M2: M2 = (0.0105 M * 0.010 L) / 0.5 L ≈ 2.1 × 10⁻⁴ M.

Find [OH⁻] concentration

Since Ca(OH)₂ produces two moles of OH⁻ ions for every mole of Ca(OH)₂, our [OH⁻] = 2 * [Ca(OH)₂] = 2 * 2.1 × 10⁻⁴ M ≈ 4.2 × 10⁻⁴ M.

Calculate the pH

We can convert the [OH⁻] to pH using the formula: pH = -log[OH⁻] = -log(4.2 × 10⁻⁴) ≈ -3.38. #d.# Calculate [OH⁻] and pH for 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

Calculate the moles of each base

Moles of Ba(OH)₂ = (0.015 M) * (0.020 L) = 3.0 × 10⁻⁴ mol. Moles of NaOH = (8.2 × 10⁻³ M) * (0.040 L) ≈ 3.3 × 10⁻⁴ mol.

Calculate the total number of moles of OH⁻

Moles of OH⁻ from Ba(OH)₂ = 2 * moles of Ba(OH)₂ = 2 * (3.0 × 10⁻⁴ mol) = 6.0 × 10⁻⁴ mol. Moles of OH⁻ from NaOH = moles of NaOH = 3.3 × 10⁻⁴ mol. Total moles of OH⁻ = 6.0 × 10⁻⁴ mol + 3.3 × 10⁻⁴ mol = 9.3 × 10⁻⁴ mol.

Calculate the total concentration of OH⁻

Total volume = 20.0 mL + 40.0 mL = 60.0 mL = 0.060 L. Total [OH⁻] = (total moles of OH⁻) / (total volume) = (9.3 × 10⁻⁴ mol) / (0.060 L) ≈ 1.55 × 10⁻² M.

Calculate the pH

We can convert the [OH⁻] to pH using the formula: pH = -log[OH⁻] = -log(1.55 × 10⁻²) ≈ -1.81.

Key concepts

Hydroxide Ion Concentration

Understanding hydroxide ion concentration is important when discussing strong base solutions. Hydroxide ions, represented as \( \text{OH}^- \), are produced by bases that dissolve completely in water. This concentration tells us how many \( \text{OH}^- \) ions are present in a solution.
A strong base like KOH dissociates completely, meaning the concentration of \( \text{OH}^- \) is equal to the concentration of the base itself. For example, if you have a 1 M KOH solution, it also has a 1 M \( \text{OH}^- \) concentration.
In cases where you have a base that can produce more than one hydroxide ion, like \( \text{Ca(OH)}_2 \), the \( \text{OH}^- \) concentration will be twice that of the initial base concentration. This is because each molecule of \( \text{Ca(OH)}_2 \) gives off two hydroxide ions as it dissolves.

pH Calculation

Calculating pH from the hydroxide ion concentration allows us to understand the acidity or basicity of a solution on a scale. The pH scale generally ranges from 0 to 14, but negative pH values can occur in instances of very strong bases or acids.
For solutions where you know \( [\text{OH}^-] \), you can find the pH by using the formula: \[ \text{pH} = 14 - \text{pOH} \], where \( \text{pOH} = -\log[\text{OH}^-] \).
This relationship comes from the fact that \( \text{pH} + \text{pOH} = 14 \) in aqueous solutions at 25°C. For example, if \( [\text{OH}^-] = 0.182 \) M, then \[ \text{pOH} = -\log(0.182) \approx 0.74 \]
and therefore \( \text{pH} = 14 - 0.74 \approx 13.26 \).
Remember that a higher pH indicates more basic conditions, with 7 being neutral.

Molarity

Molarity, a measure of concentration, is defined as the number of moles of a solute per liter of solution \((M = \frac{\text{moles}}{\text{liter}})\). It is a key concept when discussing the concentration of solutions as it helps determine how strong or weak a solution is.
To find molarity, you often start with the mass of the solute, convert that to moles using the molar mass, and then divide by the volume of the solution in liters. For example, with 3.165 grams of KOH, the molar mass is 56.11 g/mol, and this gives you about 0.0564 moles.
Placing this in 500 mL of water (or 0.5 L) results in a molarity of \( \approx 0.113 \) M for the solution, indicating a relatively strong concentration of KOH.

Dilution

Dilution refers to the process of decreasing the concentration of a solute in a solution, typically by adding more solvent. This concept is crucial when adjusting solutions to achieve desired concentrations.
When diluting solutions, the formula \( M_1V_1 = M_2V_2 \) comes in handy, where \( M_1 \) and \( V_1 \) are the initial molarity and volume, while \( M_2 \) and \( V_2 \) are the final molarity and volume. This equation tells us how the concentration changes with volume.
For instance, diluting 10 mL of a 0.0105 M \( \text{Ca(OH)}_2 \) solution to 500 mL changes its molarity to \( 2.1 \times 10^{-4} \) M. This is crucial in experiments where precise concentrations are necessary for reactions or testing.