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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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Key Concepts

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

KOH dissociation
Potassium hydroxide (KOH) is known to be a strong base. This means that when it is dissolved in water, it dissociates completely.
This dissociation follows the chemical equation: \[ \text{KOH}_{(aq)} \rightarrow \text{K}^+_{(aq)} + \text{OH}^-_{(aq)} \]Due to this complete dissociation, in an aqueous solution of KOH, the concentration of hydroxide ions—\(\text{OH}^-\)—will be the same as the initial concentration of KOH. For a 0.015 M solution of KOH:
  • The major species present are \(\text{K}^+\) and \(\text{OH}^-\) ions.
  • The concentration of \(\text{OH}^-\) ions will be 0.015 M.
Understanding KOH dissociation helps in calculating the hydroxide ion concentration and subsequently the pH of the solution.
Ba(OH)₂ dissociation
Barium hydroxide, like potassium hydroxide, is a strong base. Hence, Ba(OH)₂ also dissociates completely in water upon dissolution.
The dissociation is represented by the equation:\[ \text{Ba(OH)}_2{(aq)} \rightarrow \text{Ba}^{2+}_{(aq)} + 2\text{OH}^-_{(aq)} \]The key here is the stoichiometry of the dissociation. For every molecule of Ba(OH)₂, it produces two hydroxide ions. So, in a 0.015 M solution:
  • The major ions present are \(\text{Ba}^{2+}\) and \(\text{OH}^-\).
  • The concentration of \(\text{OH}^-\) is double that of Ba(OH)₂, hence it is 0.030 M.
This doubling effect due to the stoichiometry is crucial for determining the hydroxide ion concentration in the solution.
Hydroxide ion concentration
The concentration of hydroxide ions \([\text{OH}^-]\) is a pivotal factor in understanding the strength of a basic solution.
In strong base solutions, such as KOH and Ba(OH)₂, this concentration is directly linked to the extent of the base’s dissociation.
  • For KOH: The dissociation produces \([\text{OH}^-]\) = 0.015 M, matching the initial base concentration.
  • For Ba(OH)₂: Due to stoichiometry, two \(\text{OH}^-\) ions result in \([\text{OH}^-]\) = 0.030 M, double of its initial concentration.
Understanding and calculating these concentrations is fundamental, paving the way for computing pOH and pH values.
pOH calculation
Calculating the pOH of a solution involves using the known concentration of hydroxide ions. The formula for pOH is:
\[ \text{pOH} = -\log([\text{OH}^-]) \]Using this formula:
  • For the KOH solution, where \([\text{OH}^-] = 0.015 \text{ M}\), \[ \text{pOH} = -\log(0.015) \]
  • For the Ba(OH)₂ solution having \([\text{OH}^-] = 0.030 \text{ M}\), \[ \text{pOH} = -\log(0.030) \]
This calculation is crucial as it offers a straightforward measure of a solution's basicity. Knowing the pOH lets us determine the pH, providing a fuller picture of the solution's acidity or basicity.
pH calculation
The relationship between pH and pOH in a solution is mathematically defined as:\[ \text{pH} = 14 - \text{pOH} \]This equation derives from the auto-ionization of water. Calculating the pH:
  • First, calculate the pOH using the hydroxide ion concentration.
  • Then use the equation to determine the pH.
    For example:
    • In the KOH solution, \[ \text{pH} = 14 - \text{pOH}_{\text{KOH}} \]
    • And in the Ba(OH)₂ solution, \[ \text{pH} = 14 - \text{pOH}_{\text{Ba(OH)₂}} \]
    Knowing the pH gives a complete understanding of the solution's acidity or basicity, which is essential for many chemical applications and experiments.

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Most popular questions from this chapter

Consider a \(0.67-M\) solution of \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{2}\left(K_{\mathrm{b}}=5.6 \times 10^{-4}\right)\). a. Which of the following are major species in the solution? i. \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{2}\) ii. \(\mathrm{H}^{+}\) iii. \(\mathrm{OH}^{-}\) iv. \(\mathrm{H}_{2} \mathrm{O}\) v. \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{3}{ }^{+}\) b. Calculate the \(\mathrm{pH}\) of this solution.

Place the species in each of the following groups in order of increasing acid strength. a. \(\mathrm{H}_{2} \mathrm{O}, \mathrm{H}_{2} \mathrm{~S}, \mathrm{H}_{2} \mathrm{Se}\) (bond energies: \(\mathrm{H}-\mathrm{O}, 467 \mathrm{~kJ} / \mathrm{mol}\); \(\mathrm{H}-\mathrm{S}, 363 \mathrm{~kJ} / \mathrm{mol} ; \mathrm{H}-\mathrm{Se}, 276 \mathrm{~kJ} / \mathrm{mol})\) b. \(\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}, \mathrm{FCH}_{2} \mathrm{CO}_{2} \mathrm{H}, \mathrm{F}_{2} \mathrm{CHCO}_{2} \mathrm{H}, \mathrm{F}_{3} \mathrm{CCO}_{2} \mathrm{H}\) c. \(\mathrm{NH}_{4}^{+}, \mathrm{HONH}_{3}^{+}\) d. \(\mathrm{NH}_{4}^{+}, \mathrm{PH}_{4}{ }^{+}\) (bond energies: \(\mathrm{N}-\mathrm{H}, 391 \mathrm{~kJ} / \mathrm{mol} ; \mathrm{P}-\mathrm{H}\), \(322 \mathrm{~kJ} / \mathrm{mol}\) ) Give reasons for the orders you chose.

Calculate the \(\mathrm{pH}\) of a \(0.20-M \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{2}\) solution \(\left(K_{\mathrm{b}}=\right.\) \(\left.5.6 \times 10^{-4}\right)\)

Isocyanic acid (HNCO) can be prepared by heating sodium cyanate in the presence of solid oxalic acid according to the equation \(2 \mathrm{NaOCN}(s)+\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}(s) \longrightarrow 2 \mathrm{HNCO}(l)+\mathrm{Na}_{2} \mathrm{C}_{2} \mathrm{O}_{4}(s)\).

What mass of \(\mathrm{KOH}\) is necessary to prepare \(800.0 \mathrm{~mL}\) of a solution having a \(\mathrm{pH}=11.56\) ?

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