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Chapter 14: Problem 91

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

Short Answer

Expert verified
The major species in a 0.015 M solution of KOH are K鈦� and OH鈦� with an \(\left[ \mathrm{OH}^{-} \right]\) of 0.015 M and a pH of 12.18. The major species in a 0.015 M solution of Ba(OH)鈧� are Ba虏鈦� and OH鈦� with an \(\left[ \mathrm{OH}^{-} \right]\) of 0.030 M and a pH of 12.48.

Step by step solution

01

Identifying the major species

First, we need to identify the major species present in the solutions of the given bases. The major species are the components that exist in the solution after dissociation. a. KOH Upon dissociation, KOH will release OH鈦� ions: \[ \mathrm{KOH} \rightarrow \mathrm{K}^{+} + \mathrm{OH}^{-} \] So the major species in the solution of KOH are K鈦� and OH鈦�. b. Ba(OH)鈧� Upon dissociation, Ba(OH)鈧� releases two OH鈦� ions: \[ \mathrm{Ba(OH)_{2}} \rightarrow \mathrm{Ba}^{2+} + 2\mathrm{OH}^{-} \] So the major species in the solution of Ba(OH)鈧� are Ba虏鈦� and OH鈦�.
02

Calculating the concentration of hydroxide ions

Next, we need to calculate the concentration of hydroxide ions present in each solution. For this, we must consider the stoichiometry of the dissociation. a. KOH Since one mole of KOH dissociates to produce one mole of OH鈦�, the concentration of OH鈦� is the same as the concentration of KOH: \[ \left[ \mathrm{OH}^{-} \right] = 0.015 \,M \] b. Ba(OH)鈧� Since one mole of Ba(OH)鈧� dissociates to produce two moles of OH鈦�, the concentration of OH鈦� is twice the concentration of Ba(OH)鈧�: \[ \left[ \mathrm{OH}^{-} \right] = 2 \times 0.015 \,M = 0.030 \,M \]
03

Calculating the pH of each solution

To find the pH of each solution, we first need to calculate the pOH, which is the negative logarithm (base 10) of the hydroxide concentration: \[ \mathrm{pOH} = -\log_{10} \left[\mathrm{OH}^{-}\right] \] Then, using the relationship between pH and pOH: \[ \mathrm{pH + pOH} = 14\] We can calculate the pH of each solution: a. KOH Calculating the pOH: \[ \mathrm{pOH} = -\log_{10} \left(0.015\right) = 1.82 \] Calculating the pH: \[ \mathrm{pH} = 14 - \mathrm{pOH} = 14 - 1.82 = 12.18 \] b. Ba(OH)鈧� Calculating the pOH: \[ \mathrm{pOH} = -\log_{10} \left(0.030\right) = 1.52 \] Calculating the pH: \[ \mathrm{pH} = 14 - \mathrm{pOH} = 14 - 1.52 = 12.48 \] In summary, the major species in a solution of KOH are K鈦� and OH鈦� with an \(\left[ \mathrm{OH}^{-} \right]\) of 0.015 M and a pH of 12.18. The major species in a solution of Ba(OH)鈧� are Ba虏鈦� and OH鈦� with an \(\left[ \mathrm{OH}^{-} \right]\) of 0.030 M and a pH of 12.48.

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Key Concepts

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

Acid-Base Reactions
Acid-base reactions are a fundamental part of chemistry and involve the transfer of protons (H鈦� ions). Bases are substances that accept protons, and when they dissociate in water, they produce hydroxide ions (OH鈦�). This makes them distinct from acids, which donate protons. Bases are usually compounds containing OH groups, such as potassium hydroxide (KOH) or barium hydroxide (Ba(OH)鈧�). When these bases are introduced into water, they dissociate and release OH鈦� ions. The nature of this dissociation is what classifies these compounds as bases. In our discussion, we are primarily concerned with how these ions react in the water to establish equilibrium, particularly involving the hydroxide ions produced in the reactions.
pH Calculation
pH calculation is an essential part of understanding the acidity or basicity of a solution. pH is a measure of the concentration of hydrogen ions (H鈦�) in a solution, with a low pH indicating an acidic solution and a high pH indicating a basic one. In solutions of bases like KOH and Ba(OH)鈧�, we actually calculate the pOH first by using the concentration of OH鈦� ions:
  • pOH is computed as the negative logarithm (base 10) of the hydroxide ion concentration, represented as \[-\log_{10} \left[\mathrm{OH}^{-}\right]\].
  • To convert pOH to pH, use the relation \[\mathrm{pH} + \mathrm{pOH} = 14\]. This relation holds true as pH and pOH are complements with respect to the ion product of water (14 at 25掳C).
Calculating the pH of these basic solutions involves first determining the hydroxide ion concentration, then using the steps above to find pOH and finally converting this to pH.
Dissociation of Bases
The dissociation of bases is a key concept in chemical equilibria as it directly affects the pH of the solution. Strong bases like KOH and Ba(OH)鈧� dissociate completely in water. This means that when KOH is dissolved, it fully splits into K鈦� and OH鈦� ions. Meanwhile, Ba(OH)鈧� dissociates into Ba虏鈦� and two OH鈦� ions.
  • For KOH, each mole of the substance yields one mole of OH鈦�, simply reflecting its 1:1 dissociation \[[\mathrm{KOH} \rightarrow \mathrm{K}^{+} + \mathrm{OH}^{-}]\].
  • Ba(OH)鈧�, however, dissociates to give twice the amount of OH鈦� ions per mole \[[\mathrm{Ba(OH)_{2}} \rightarrow \mathrm{Ba}^{2+} + 2 \mathrm{OH}^{-}]\].
Understanding the stoichiometry of these reactions allows us to accurately compute the concentration of OH鈦� in solution, which is pivotal in calculating the resulting pH and determining the solution's basicity.

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Most popular questions from this chapter

At \(25^{\circ} \mathrm{C},\) a saturated solution of benzoic acid \(\left(K_{\mathrm{a}}=6.4 \times 10^{-5}\right)\) has a pH of \(2.80 .\) Calculate the water solubility of benzoic acid in moles per liter.

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, \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{NH}\)

Calculate the pH of a \(0.010-M\) solution of iodic acid (HIO \(_{3}, K_{\mathrm{a}}\) \(=0.17 )\)

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Aluminum hydroxide is an amphoteric substance. It can act as either a Bronsted-Lowry base or a Lewis acid. Write a reaction showing All(OH) acting as a base toward \(\mathrm{H}^{+}\) and as an acid toward OH".
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