Question

Calculate \(\left[\mathrm{OH}^{-}\right]\) and \(\mathrm{pH}\) for each of the following strong base solutions: (a) \(0.082 \mathrm{M} \mathrm{KOH}\), (b) \(1.065 \mathrm{~g}\) 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 \(7.5 \times 10^{-3} \mathrm{M} \mathrm{NaOH}\)

Short answer

(a) The hydroxide ion concentration, \[\left[\mathrm{OH}^{-}\right] = 0.082 \mathrm{M}\]. To find the pH, first calculate pOH, then pH as \(\mathrm{pH} = 14 - \mathrm{pOH}\). (b) Convert the mass of KOH to moles, then calculate the concentration of OH⁻ ions by dividing the moles of KOH by the volume of the solution. Finally, calculate pOH and then pH as in solution (a). (c) Calculate the moles of OH⁻ ions provided by Ca(OH)₂, and then use the final volume of the diluted solution to calculate the concentration of OH⁻ ions. Calculate pOH and then pH as in solution (a). (d) Calculate the moles of OH⁻ ions provided by NaOH, and then use the total volume of the solution to determine the concentration of OH⁻ ions. Calculate pOH and then pH as in solution (a).

Step-by-step solution

Calculate the moles of OH- ions

In a 0.082 M KOH solution, the concentration of OH⁻ ions is the same since each KOH molecule dissociates into one OH⁻ ion. Therefore, \[\left[\mathrm{OH}^{-}\right] = 0.082 \mathrm{M}\]

Calculate pH

To calculate the pH, we first have to determine the pOH. \[\mathrm{pOH} = -\log \left[\mathrm{OH}^{-}\right]\] Then, \[\mathrm{pH} = 14 - \mathrm{pOH}\] (b) Step 1: Convert mass of KOH to moles of OH⁻ ions

Convert mass of KOH to moles

Given the mass of KOH is 1.065 g. The molar mass of KOH is 56.11 g/mol. Moles of KOH = \[\frac{1.065\,\mathrm{g}}{56.11\,\mathrm{g/mol}}\]

Calculate the concentration of OH⁻ ions

Divide the moles of KOH by the volume of the solution to get the concentration of OH⁻ ions. \[\left[\mathrm{OH}^{-}\right] = \frac{\text{moles of KOH}}{0.500\,\mathrm{L}}\]

Calculate pH

Calculate pOH and then pH as in solution (a). (c) Step 1: Calculate the moles of OH⁻ ions in 10.0 mL of 0.0105 M Ca(OH)₂

Calculate the moles of OH⁻ ions

One molecule of Ca(OH)₂ provides 2 moles of OH⁻ ions. So, Moles of OH⁻ ions = 2 x (0.0105 M × 0.010 L)

Calculate the concentration of OH⁻ ions

Divide the moles of OH⁻ ions by the final volume of the diluted solution. \[\left[\mathrm{OH}^{-}\right] = \frac{\text{moles of OH}^{-}}{0.500\,\mathrm{L}}\]

Calculate pH

Calculate pOH and then pH as in solution (a). (d) Step 1: Calculate the moles of OH⁻ ions

Calculate the moles of OH⁻ ions

In this solution, one molecule of NaOH provides one mole of OH⁻ ions. Moles of OH⁻ ions = 7.5 × 10⁻³ M × (volume of NaOH solution in L)

Determine the concentration of OH⁻ ions

Assume the volume of NaOH solution is small compared to the total volume, and use 500 mL as the final volume for the calculation of \[\left[\mathrm{OH}^{-}\right]\]. \[\left[\mathrm{OH}^{-}\right] = \frac{\text{moles of OH}^{-}}{0.500\,\mathrm{L}}\]

Calculate pH

Calculate pOH and then pH as in solution (a).

Key concepts

Hydroxide Ion Concentration

One of the most fundamental concepts in understanding strong base solutions is the hydroxide ion concentration, denoted as \( [OH^-] \). This measurement dictates the strength of a base in solution. For example, in a solution of potassium hydroxide (KOH), each molecule fully dissociates to produce one hydroxide ion. Therefore, a \( 0.082 \mathrm{M} \) KOH solution has an \( [OH^-]\) of \( 0.082 \mathrm{M} \).

For a given mass of KOH, we can calculate \( [OH^-]\) by first finding the moles of KOH using its molar mass. With \( 1.065 \mathrm{g} \) of KOH and a molar mass of \( 56.11 \mathrm{g/mol} \), you would divide the grams by the molar mass to find the moles, and then divide by the volume in liters to determine \( [OH^-]\).

In the case of calcium hydroxide, \( \mathrm{Ca(OH)_2}\), each formula unit yields two hydroxide ions when dissolved. So if you have a \( 0.0105 \mathrm{M} \) solution, in every liter of solution there would be \( 0.0210 \mathrm{M} \) hydroxide ions because of this stoichiometry.

pOH and pH Relationship

Understanding the pOH and pH relationship is crucial for mastering acid-base chemistry. The pOH is a measure of the hydroxide ion concentration and is calculated using the formula \( pOH = -\log[OH^-]\). The pH, which measures the acidity of a solution, can then be found since pH and pOH are inversely related in a neutral solution at 25°C. Their sum is always 14, so \( pH = 14 - pOH\).

Using the formula, you can calculate the pH of any strong base solution once you know the concentration of hydroxide ions. For instance, if you have a pOH of 1.2 for a certain base solution, the pH would be calculated as \( 14 - 1.2 = 12.8\), indicating a strongly basic solution.

Molarity and Dilution

Calculating the molarity, or molar concentration, of a solution is essential when working with chemical solutions. Molarity is defined as moles of solute per liter of solution (\( M = \frac{moles}{L} \) ). Dilution plays a significant role in determining the final concentration of a solution. When a solution is diluted, its volume increases, but the number of moles of solute remains constant.

For example, if you dilute \( 10.0 \mathrm{mL} \) of a \( 0.0105 \mathrm{M} \) \( \mathrm{Ca(OH)_2}\) solution to \( 500.0 \mathrm{mL}\), you still have the same number of moles of \( \mathrm{Ca(OH)_2}\) as you started with. But because the volume is now \( 500.0 \mathrm{mL}\), or \( 0.5 \mathrm{L}\), the concentration will change. To find the new molarity after dilution, you divide the original moles of solute by the new volume in liters.

Stoichiometry of Base Dissociation

Stoichiometry is the part of chemistry that deals with the quantitative relationships that govern the formation of substances in chemical reactions. For bases, this involves looking at how the base molecules dissociate in water to produce hydroxide ions. The stoichiometry of base dissociation refers to the ratio of hydroxide ions produced per formula unit of the base.

In the dissociation process, a base like \( \mathrm{KOH}\) will produce one hydroxide ion per molecule. However, a base like \( \mathrm{Ca(OH)_2}\) produces two hydroxide ions for each formula unit that dissolves. It’s important to remember this ratio, especially when calculating concentration, because it affects the total hydroxide ion concentration. This relationship dictates how we proceed with the calculations for any base when solving for pH and other related properties.