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Chapter 14: Problem 91

What are the major species present in \(0.015 \mathrm{M}\) solutions of each of the following bases? a. \(\mathrm{KOH}\) b. \(\mathrm{Ba}(\mathrm{OH})_{2}\) What is \(\left[\mathrm{OH}^{-}\right]\) and the \(\mathrm{pH}\) of each of these solutions?

Short Answer

Expert verified
In the 0.015 M KOH solution, the major species present are K⁺ and OH⁻ ions from the complete dissociation of KOH, with [OH⁻] = 0.015 M. The pOH of the solution is -log(0.015) and the pH is calculated as 14 - pOH_KOH. In the 0.015 M Ba(OH)₂ solution, the major species present are Ba²⁺ and OH⁻ ions from the complete dissociation of Ba(OH)₂, with [OH⁻] = 0.030 M. The pOH of the solution is -log(0.030) and the pH is calculated as 14 - pOH_Ba(OH)2.

Step by step solution

01

Potassium hydroxide (KOH) is a strong base, meaning it completely dissociates in water. We can represent this dissociation with the following balanced chemical equation: KOH _(aq) -> K⁺ _(aq) + OH⁻ _(aq) #step 2: Dissociation of Ba(OH)2 in water#

Barium hydroxide (Ba(OH)₂) is also a strong base and will completely dissociate in water. The balanced chemical equation for its dissociation is: Ba(OH)₂ _(aq) -> Ba²⁺ _(aq) + 2OH⁻ _(aq) #step 3: Major species present in the KOH solution#
02

In the 0.015 M KOH solution, the major species present are K⁺ and OH⁻ ions from the complete dissociation of KOH. Since the stoichiometry is 1:1, the concentration of each ion will also be 0.015 M. #step 4: Major species present in the Ba(OH)2 solution#

In the 0.015 M Ba(OH)₂ solution, the major species present are Ba²⁺ and OH⁻ ions from the complete dissociation of Ba(OH)₂. However, notice that the stoichiometry for the dissociation of barium hydroxide is 1:2, meaning that the concentration of OH⁻ ions will be twice that of Ba²⁺ ions. Thus, the concentration of OH⁻ in this solution will be 2 * 0.015 M = 0.030 M. #step 5:Calculating [OH⁻] and pOH for each solution#
03

In the KOH solution, [OH⁻] = 0.015 M. To find the pOH of the solution, we use the formula pOH = -log([OH⁻]). For the KOH solution: pOH = -log(0.015) In the Ba(OH)₂ solution, [OH⁻] = 0.030 M. Similarly, to find the pOH of the solution: pOH = -log(0.030 ) #step 6:Calculating pH for each solution#

Finally, we'll use the relationship between pH and pOH to calculate the pH of each solution. This is given by the equation pH = 14 - pOH. For the KOH solution: pH = 14 - pOH_KOH For the Ba(OH)₂ solution: pH = 14 - pOH_Ba(OH)2 After completing calculations for the pOH and pH values, we will have our final answers for the concentrations [OH⁻], pOH, and pH of each solution.

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Most popular questions from this chapter

Given that the \(K_{\mathrm{a}}\) value for acetic acid is \(1.8 \times 10^{-5}\) and the \(K_{\text {a }}\) value for hypochlorous acid is \(3.5 \times 10^{-8}\), which is the stronger base, \(\mathrm{OCl}^{-}\) or \(\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{O}_{2}^{-}\) ?

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