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Chapter 14: Problem 92

What are the major species present in the following mixtures of bases? a. 0.050 M NaOH and 0.050 M LiOH b. 0.0010\(M \mathrm{Ca}(\mathrm{OH})_{2}\) and 0.020\(M \mathrm{RbOH}\) What is [OH \(^{-} ]\) and the pH of each of these solutions?

Short Answer

Expert verified
For the given solutions: (a): Major species: NaOH, LiOH [OH⁻] = 0.100 M pH = 13 (b): Major species: Ca(OH)₂, RbOH [OH⁻] = 0.0220 M pH ≈ 12.34

Step by step solution

01

(Step 1 - Identify Major Species)

In both (a) and (b), it's clear that the major species present are NaOH, LiOH, Ca(OH)₂, and RbOH.
02

(Step 2 - Calculate Total [OH⁻])

To calculate the total concentration of OH⁻ in both the solutions, we simply add the contributed amount of hydroxide ions from each of the bases present in the solution. (a) 0.050 M NaOH and 0.050 M LiOH For NaOH and LiOH: [OH⁻] = [OH⁻ from NaOH] + [OH⁻ from LiOH] = 0.050 M + 0.050 M = 0.100 M (b) 0.0010 M Ca(OH)₂ and 0.020 M RbOH In this solution, we must account for the fact that Ca(OH)₂ releases two OH⁻ ions per molecule. For RbOH: [OH⁻] = 0.020 M For Ca(OH)₂: [OH⁻] = 2 * 0.0010 M = 0.0020 M Total [OH⁻] = [OH⁻ from RbOH] + [OH⁻ from Ca(OH)₂] = 0.020 M + 0.0020 M = 0.0220 M
03

(Step 3 - Calculate pOH)

Now, we need to calculate the pOH of each solution using the formula: pOH = -log [OH⁻] For (a) solution: pOH = -log(0.100) = 1 For (b) solution: pOH = -log(0.0220) ≈ 1.66
04

(Step 4 - Calculate pH)

Finally, we calculate the pH of each solution using the formula: pH = 14 - pOH For (a) solution: pH = 14 - 1 = 13 For (b) solution: pH ≈ 14 - 1.66 ≈ 12.34 So for the given solutions, we have the following results: (a): Major species: NaOH, LiOH [OH⁻] = 0.100 M pH = 13 (b): Major species: Ca(OH)₂, RbOH [OH⁻] = 0.0220 M pH ≈ 12.34

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