Question

Base Solutions For each strong base solution, determine \(\left[\mathrm{OH}^{-}\right],\left[\mathrm{H}_{3} \mathrm{O}^{+}\right], \mathrm{pH},\) and \(\mathrm{pOH} .\) $$\begin{array}{l}{\text { a. } 8.77 \times 10^{-3} \mathrm{M} \text { LiOH }} \\\ {\text { b. } 0.0112 \mathrm{M} \mathrm{Ba}(\mathrm{OH})_{2}} \\ {\text { c. } 1.9 \times 10^{-4} \mathrm{MKOH}_{2}} \\ {\text { d. } 5.0 \times 10^{-4} \mathrm{M} \mathrm{Ca}(\mathrm{OH})_{2}}\end{array}$$

Short answer

a. \([\mathrm{OH}^-] = 8.77 \times 10^{-3}\), \([\mathrm{H}_{3}\mathrm{O}^+] = 1.14 \times 10^{-12}\), pH = 11.94, pOH = 2.06. b. \([\mathrm{OH}^-] = 0.0224\), \([\mathrm{H}_{3}\mathrm{O}^+] = 4.46 \times 10^{-13}\), pH = 12.35, pOH = 1.65. c. \([\mathrm{OH}^-] = 1.9 \times 10^{-4}\), \([\mathrm{H}_{3}\mathrm{O}^+] = 5.26 \times 10^{-11}\), pH = 10.28, pOH = 3.72. d. \([\mathrm{OH}^-] = 1.0 \times 10^{-3}\), \([\mathrm{H}_{3}\mathrm{O}^+] = 1.0 \times 10^{-11}\), pH = 11.00, pOH = 3.00.

Step-by-step solution

Understanding Strong Bases

Recognize that strong bases completely dissociate in water. For monobasic bases like LiOH and KOH, the concentration of the base equals the concentration of \(\mathrm{OH}^-\). For dibasic bases like \(\mathrm{Ba}(\mathrm{OH})_2\) and \(\mathrm{Ca}(\mathrm{OH})_2\), each base molecule produces two \(\mathrm{OH}^-\) ions upon dissociation.

Calculating \(\left[\mathrm{OH}^-\right]\)

For (a) LiOH and (c) KOH, the concentration of \(\mathrm{OH}^-\) is the same as the given concentration of the base. For (b) \(\mathrm{Ba}(\mathrm{OH})_2\) and (d) \(\mathrm{Ca}(\mathrm{OH})_2\), multiply the given concentration of the base by 2 to get \(\left[\mathrm{OH}^-\right]\).

Calculating \(\left[\mathrm{H}_3\mathrm{O}^+\right]\) and \(\mathrm{pH}\)

Apply the water dissociation constant \(K_w = 1.0 \times 10^{-14}\) at 25 degrees Celsius to find \(\left[\mathrm{H}_3\mathrm{O}^+\right]\) using the formula \(K_w = \left[\mathrm{OH}^-\right] \times \left[\mathrm{H}_3\mathrm{O}^+\right]\). Then, calculate pH as \(\mathrm{pH} = -\log(\left[\mathrm{H}_3\mathrm{O}^+\right])\).

Calculating \(\mathrm{pOH}\)

Use the relationship \(\mathrm{pOH} = -\log(\left[\mathrm{OH}^-\right])\). With pH and pOH known, recall that \(\mathrm{pH} + \mathrm{pOH} = 14\) to verify calculations are consistent.

Determine Values for Each Base Solution

For each base, perform the calculations following the previous steps. Remember to account for whether the base is monobasic or dibasic in the calculations.

Key concepts

Hydroxide Ion Concentration

When we talk about hydroxide ion concentration \(\left[\mathrm{OH}^-\right]\), we are referring to the number of hydroxide ions present in a solution. In the context of strong bases, this is directly related to the dissociation of the base in water. Strong bases, such as lithium hydroxide (LiOH) and potassium hydroxide (KOH), dissociate completely, meaning each molecule of the strong base gives rise to one hydroxide ion for monobasic (single hydroxide ion) bases in solution.

For dibasic bases like barium hydroxide (\(\mathrm{Ba}(\mathrm{OH})_2\)) or calcium hydroxide (\(\mathrm{Ca}(\mathrm{OH})_2\)), which contain two hydroxide ions per molecule, the concentration of hydroxide ions will be twice the concentration of the base itself. Thus, to calculate the \(\left[\mathrm{OH}^-\right]\) for these solutions, you'll often multiply the initial concentration value by the number of hydroxide ions that each base molecule can produce.

Hydronium Ion Concentration

On the flip side, the hydronium ion concentration \(\left[\mathrm{H}_3\mathrm{O}^+\right]\) measures the number of hydronium ions in solution. For an aqueous solution, the product of hydronium ion concentration and hydroxide ion concentration is always constant at \(10^{-14}\) at 25 degrees Celsius, based on the water dissociation constant \(K_w\).

In a solution with a known hydroxide ion concentration, you can find the hydronium ion concentration by dividing \(K_w\) by \(\left[\mathrm{OH}^-\right]\). This is particularly useful in strong base solutions because, with complete dissociation, determining the concentration of one component immediately allows for the calculation of the other. Remember the fundamental relationship: \(K_w = \left[\mathrm{OH}^-\right] \times \left[\mathrm{H}_3\mathrm{O}^+\right]\), which remains a cornerstone of acid-base chemistry.

pH and pOH Relationship

Understanding the pH and pOH relationship is crucial for mastering acid-base chemistry. pH is a measure of the acidity of a solution, whereas pOH is a measure of the basicity. Both scales are logarithmic, inversely related to the concentration of hydronium ions and hydroxide ions respectively. The pH scale typically ranges from 0 to 14, with 7 being neutral, lower values being acidic, and higher values being basic.

The relationship between pH and pOH can be summarized by the equation \(\text{pH} + \text{pOH} = 14\), which holds true for all aqueous solutions at 25 degrees Celsius. If you know the pH, you can calculate the pOH, and vice versa. To find pH, use the formula \(\text{pH} = -\log(\left[\mathrm{H}_3\mathrm{O}^+\right])\) and for pOH, \(\text{pOH} = -\log(\left[\mathrm{OH}^-\right])\). This relationship is especially helpful when dealing with strong bases because once \(\left[\mathrm{OH}^-\right]\) is known, one can readily calculate the pOH and subsequently the pH.

Dissociation of Strong Bases

Dissociation is the process through which ionic compounds separate into their ions when dissolved in water. With strong bases, dissociation is a key concept because they fully dissociate into their respective ions. This complete dissociation means that no matter the initial concentration of the strong base, you can expect a stoichiometrically equivalent concentration of hydroxide ions in solution.

Dissociation is an all-or-nothing event for strong bases. For example, the monobasic base LiOH in solution is entirely present as Li+ and OH- ions. In contrast, each molecule of the dibasic base \(\mathrm{Ba}(\mathrm{OH})_2\) or \(\mathrm{Ca}(\mathrm{OH})_2\) dissociates into one Ba2+ or one Ca2+ ion and two OH- ions. This complete dissociation simplifies calculations involving strong bases, ensuring a direct relationship between the initial concentration of the base and the concentration of hydroxide ions, which is exactly what's needed for the pH, pOH, and other related calculations.