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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鈦籡 = 0.100 M pH = 13 (b): Major species: Ca(OH)鈧, RbOH [OH鈦籡 = 0.0220 M pH 鈮 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)鈧, and RbOH.
02

(Step 2 - Calculate Total [OH鈦籡)

To calculate the total concentration of OH鈦 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鈦籡 = [OH鈦 from NaOH] + [OH鈦 from LiOH] = 0.050 M + 0.050 M = 0.100 M (b) 0.0010 M Ca(OH)鈧 and 0.020 M RbOH In this solution, we must account for the fact that Ca(OH)鈧 releases two OH鈦 ions per molecule. For RbOH: [OH鈦籡 = 0.020 M For Ca(OH)鈧: [OH鈦籡 = 2 * 0.0010 M = 0.0020 M Total [OH鈦籡 = [OH鈦 from RbOH] + [OH鈦 from Ca(OH)鈧俔 = 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鈦籡 For (a) solution: pOH = -log(0.100) = 1 For (b) solution: pOH = -log(0.0220) 鈮 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 鈮 14 - 1.66 鈮 12.34 So for the given solutions, we have the following results: (a): Major species: NaOH, LiOH [OH鈦籡 = 0.100 M pH = 13 (b): Major species: Ca(OH)鈧, RbOH [OH鈦籡 = 0.0220 M pH 鈮 12.34

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Hydroxide Ion Concentration
In base mixtures, understanding hydroxide ion concentration is crucial for determining the pH of a solution. Hydroxide ions, denoted by OH鈦, are typically released by bases when dissolved in water. The concentration of these ions directly affects the basicity, and consequently, the pH of the solution.
For instance, a solution containing 0.050 M NaOH and 0.050 M LiOH releases hydroxide ions from both sodium hydroxide and lithium hydroxide. Each 1 mole of both NaOH and LiOH releases one mole of OH鈦. Therefore, by adding their contributions, we get the total concentration of OH鈦 as 0.050 M + 0.050 M = 0.100 M.
For a second mixture, with 0.0010 M Ca(OH)鈧 and 0.020 M RbOH, calcium hydroxide contributes more due to its ability to release two OH鈦 ions per molecule. So, the concentration from Ca(OH)鈧 is 2 * 0.0010 M = 0.0020 M, and for RbOH, it is 0.020 M. The total hydroxide concentration is 0.0020 M + 0.020 M = 0.0220 M.
Major Species in Solutions
Identifying the major species in a base solution helps understand the solution's behavior and reactivity. The term 'major species' refers to the dominant ions and molecules present when bases dissolve in water.
In mixtures of bases, such as NaOH with LiOH or Ca(OH)鈧 with RbOH, the major species are the compounds that primarily contribute to the solution's chemical properties. In the first mixture, with NaOH and LiOH, the major species present are the sodium ions (Na鈦), lithium ions (Li鈦), and hydroxide ions (OH鈦). These ions primarily determine the solution's characteristics.
In the second mixture containing Ca(OH)鈧 and RbOH, the major species include calcium ions (Ca虏鈦), rubidium ions (Rb鈦), and hydroxide ions (OH鈦). These ions come from the dissociation of the strong base molecules in water, ensuring the solution remains basic and dictates its overall properties.
Step by Step pH Calculation
Calculating pH from a given hydroxide ion concentration involves several simple steps. This process allows for determining the acidity or basicity of the solution effectively.
The process begins by calculating pOH using the formula:\[ \text{pOH} = -\log [\text{OH}^-] \]From our example, for the mixture of 0.100 M OH鈦, the pOH is given by:\[ \text{pOH} = -\log(0.100) = 1 \]Similarly, for the solution with 0.0220 M OH鈦, the calculation becomes:\[ \text{pOH} = -\log(0.0220) \approx 1.66 \]Finally, with the pOH known, the pH can be determined using the relationship between pH and pOH:\[ \text{pH} = 14 - \text{pOH} \]So, the pH for the first solution is:\[ \text{pH} = 14 - 1 = 13 \]And, for the second solution:\[ \text{pH} \approx 14 - 1.66 \approx 12.34 \]
Base Mixtures
In chemistry, base mixtures often contain multiple strong bases dissolved together in a solution. Understanding these mixtures is key to calculating properties like pH and OH鈦 concentration.
When mixing bases, each contributes hydroxide ions depending on their concentration in the solution. For instance, combining equal molarities of NaOH and LiOH leads to a simple addition of their hydroxide ion contributions. However, when mixing different bases, like Ca(OH)鈧 and RbOH, the stoichiometry may differ as Ca(OH)鈧 can release more hydroxide ions per molecule.
The reactivity and concentration contribute to the ease of calculating total [OH鈦籡 and consequent pH. These bases, being strong, dissociate completely, simplifying the calculation process as each base's contribution can be directly added to find the resulting hydroxide ion concentration.
Understanding the behavior of these solutions is fundamental to accurately predicting the chemical nature and environmental impact of base mixtures.

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Most popular questions from this chapter

A typical sample of vinegar has a pH of \(3.0 .\) Assuming that vinegar is only an aqueous solution of acetic acid \(\left(K_{\mathrm{a}}=1.8 \times\right.\) \(10^{-5}\) ), calculate the concentration of acetic acid in vinegar.

Identify the Lewis acid and the Lewis base in each of the following reactions. a. \(\mathrm{Fe}^{3+}(a q)+6 \mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}^{3+}(a q)\) b. \(\mathrm{H}_{2} \mathrm{O}(l)+\mathrm{CN}^{-}(a q) \Longrightarrow \mathrm{HCN}(a q)+\mathrm{OH}^{-}(a q)\) c. \(\mathrm{HgI}_{2}(s)+2 \mathrm{I}^{-}(a q) \rightleftharpoons \mathrm{Hgl}_{4}^{2-}(a q)\)

Write the reaction and the corresponding \(K_{\mathrm{b}}\) equilibrium expression for each of the following substances acting as bases in water. a. aniline, \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}\) b. dimethylamine, \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{NH}\)

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