Question

Calculate the \(\mathrm{pOH}\) and \(\mathrm{pH}\) of the following aqueous solutions at \(25^{\circ} \mathrm{C}:\) (a) \(0.066 \mathrm{M} \mathrm{KOH},\) (b) \(5.43 \mathrm{M} \mathrm{NaOH}\), (c) \(0.74 \mathrm{M} \mathrm{Ba}(\mathrm{OH})_{2}\).

Short answer

(a) pH = 12.82, (b) pH ≈ 14 (strongly basic), (c) pH ≈ 14 (strongly basic).

Step-by-step solution

Understanding the Relationship Between pH, pOH, and Hydroxide Concentration

For an aqueous solution, the pH and pOH are related by the equation: \( \text{pH} + \text{pOH} = 14 \) at \( 25^{\circ} \mathrm{C} \). Since KOH, NaOH, and \( \mathrm{Ba(OH)}_2 \) are strong bases, they dissociate completely in water. Therefore, for each mole of KOH and NaOH, the concentration of \( \mathrm{OH}^- \) ions is the same as the original concentration of the base. For \( \mathrm{Ba(OH)}_2 \), this is doubled as each unit produces two \( \mathrm{OH}^- \) ions.

Calculate the Concentration of OH- Ions

(a) For \(0.066 \mathrm{M} \mathrm{KOH}\), the \( \mathrm{OH}^- \) concentration is \(0.066 \mathrm{M}\). (b) For \(5.43 \mathrm{M} \mathrm{NaOH}\), the \( \mathrm{OH}^- \) concentration is \(5.43 \mathrm{M}\). (c) For \(0.74 \mathrm{M} \mathrm{Ba(OH)}_2\), each mole gives two moles of \( \mathrm{OH}^- \) ions, so the \( \mathrm{OH}^- \) concentration is \(2 \times 0.74 = 1.48 \mathrm{M}\).

Calculate pOH from OH- Concentration

The pOH is calculated using the formula: \( \text{pOH} = -\log[\mathrm{OH}^-] \).(a) \( \text{pOH} = -\log(0.066) = 1.18 \)(b) \( \text{pOH} = -\log(5.43) = -0.73 \)(c) \( \text{pOH} = -\log(1.48) = -0.17 \)

Calculate pH from pOH

Using the relationship \( \text{pH} + \text{pOH} = 14 \), we can find pH:(a) \( \text{pH} = 14 - 1.18 = 12.82 \)(b) \( \text{pH} = 14 + 0.73 = 14.73 \) (Note: pH cannot exceed 14, but the concept highlights the strength of the basic solution)(c) \( \text{pH} = 14 + 0.17 = 14.17 \)

Key concepts

pOH Calculation

Understanding how to calculate pOH is crucial when dealing with basic solutions. The pOH is a measure of the concentration of hydroxide ions (\([\text{OH}^-]\)) in a solution. To calculate pOH, we use the logarithmic formula:

• \( \text{pOH} = -\log[\text{OH}^-]\)

The minus sign indicates that logarithms transform the concentration of ions into a more manageable number, especially for very small or very large concentrations.
A low pOH indicates a high concentration of hydroxide ions, hence, a strong base. Conversely, a high pOH suggests a weaker base or a more acidic solution, characterized by fewer hydroxide ions. Keep in mind that the relation with pH is given by:

• \( \text{pH} + \text{pOH} = 14 \)

Strong Bases

Understanding strong bases is a key aspect of acid-base chemistry. A strong base is one that fully dissociates into its ions in solution. This means that every unit of a strong base produces hydroxide ions. Common examples include potassium hydroxide (KOH), sodium hydroxide (NaOH), and barium hydroxide (Ba(OH)\(_2\)).

• KOH and NaOH dissociate to produce one mole of \( \text{OH}^- \) ions per mole of base.
• Ba(OH)\(_2\) dissociates to yield two moles of \( \text{OH}^- \) ions per mole of base due to its two hydroxide groups.

This complete dissociation makes strong bases very effective at increasing the hydroxide ion concentration in solutions, resulting in high pH and low pOH values.

Molarity

Molarity is a measure of concentration, specifically the number of moles of solute per liter of solution. When we express the concentration of a strong base in molarity, it directly indicates how many moles of hydroxide ions are released once the base dissociates in solution.
For strong bases like KOH and NaOH, the molarity of the base equals the molarity of \(\text{OH}^-\) ions since they dissociate fully into one mole of ions per mole of base. However, for compounds like Ba(OH)\(_2\), which release two moles of \(\text{OH}^-\) ions per mole of base, the hydroxide ion concentration (\(\text{[OH]^-}\)) is effectively twice the molarity of the base.This measure becomes vital when calculating pOH and working with dilution or reaction stoichiometry.

Dissociation

In chemistry, dissociation refers to the process by which an ionic compound separates into its individual ions in a solution. When strong bases dissociate in water, the ionic bonds break, releasing hydroxide ions (\(\text{OH}^-\)) into the solution.This process is complete for strong bases, meaning that all of the base molecules dissociate fully, significantly increasing the solution's hydroxide concentration. For instance:

• KOH dissolves into \(\mathrm{K}^+\) and \(\text{OH}^-\) ions.
• NaOH splits into \(\mathrm{Na}^+\) and \(\text{OH}^-\) ions.
• Ba(OH)\(_2\) dissociates into \(\mathrm{Ba}^{2+}\) and two \(\text{OH}^-\) ions.

Understanding dissociation helps in predicting how the hydroxide ion concentration affects the pOH and pH of a solution, as it directly influences the calculations and properties of the base.