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Boric acid and glycerin form a complex \(\mathrm{B}(\mathrm{OH})_{3}(\mathrm{aq})+\) glycerin \((\mathrm{aq}) \rightleftharpoons \mathrm{B}(\mathrm{OH})_{3} \cdot\) glycerin(aq) with an equilibrium constant of 0.90. If the concentration of boric acid is 0.10 M, how much glycerin should be added, per liter, so that \(60 . \%\) of the boric acid is in the form of the complex?

Short Answer

Expert verified
Add 1.67 M of glycerin per liter.

Step by step solution

01

Understanding the Equilibrium

The given system reaches equilibrium with the reaction \( \text{B(OH)}_3 + \text{glycerin} \rightleftharpoons \text{B(OH)}_3 \cdot \text{glycerin} \). The equilibrium constant \( K \) is 0.90. We need to ensure that 60% of boric acid forms the complex.
02

Setting Up the Expression

Let the initial concentration of boric acid be \( 0.10 \ \text{M} \). If 60% forms the complex, the concentration of the complex at equilibrium is \( 0.06 \ \text{M} \). The concentration of unreacted boric acid at equilibrium is \( 0.10 - 0.06 = 0.04 \ \text{M} \).
03

Applying the Equilibrium Constant

The equilibrium constant \( K = 0.90 \) is expressed as \[ K = \frac{[\text{B(OH)}_3 \cdot \text{glycerin}]}{[\text{B(OH)}_3][\text{glycerin}]} = \frac{0.06}{0.04 \times [\text{glycerin}]} \]
04

Solving for Glycerin Concentration

Rearrange the equation and solve for the glycerin concentration: \[ 0.90 \times 0.04 \times [\text{glycerin}] = 0.06 \]\[ [\text{glycerin}] = \frac{0.06}{0.90 \times 0.04} = 1.67 \ \text{M} \]
05

Conclusion

To achieve the desired complex formation, add approximately \( 1.67 \ \text{M} \) of glycerin per liter to the solution.

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Key Concepts

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

Boric Acid
Boric acid, with the chemical formula \(\text{B(OH)}_3\), is a weak monobasic acid often used in antiseptics, insecticides, and various chemical reactions. It is known for its gentle properties and ability to form complexes with other molecules.
When added to water, boric acid does not fully dissociate, which means it remains largely in its original state. This characteristic makes it useful in forming complexes, as it can interact with other chemical species without fully breaking down.
  • Features weak acidity
  • Forms complexes with agents like glycerin
  • Used in various applications, from medical to industrial
Glycerin
Glycerin, also known as glycerol, is a simple polyol compound that is colorless, odorless, and sweet-tasting. It has three hydroxyl (\(-OH\)) groups, making it highly soluble in water and enabling it to form hydrogen bonds with other molecules.
In the context of chemical reactions, glycerin can serve as a stabilizing or complexing agent. Its ability to participate in hydrogen bonding allows it to readily interact with other compounds like boric acid.
  • Highly soluble in water due to multiple hydroxyl groups
  • Sweet-tasting and non-toxic
  • Used in cosmetics, food, and pharmaceuticals
Equilibrium Constant
The equilibrium constant \( K \) is a crucial concept in understanding chemical equilibria. It quantifies the ratio of concentrations of products to reactants at equilibrium at a given temperature. In this reaction, it is given as \( K = 0.90 \).
The equilibrium constant helps to understand how far a reaction will proceed. For a reaction:
\(\text{B(OH)}_3 + \text{glycerin} \rightleftharpoons \text{B(OH)}_3 \cdot \text{glycerin}\)
The expression for \( K \) is:
\[ K = \frac{[\text{B(OH)}_3 \cdot \text{glycerin}]}{[\text{B(OH)}_3][\text{glycerin}]} \]
A \( K \) value of 0.90 indicates that the reaction slightly favors the formation of the complex, but not overwhelmingly so.
  • Reflects the balance between reactants and products
  • Aids in calculating concentrations needed for desired outcomes
Complex Formation
Complex formation involves the combination of two or more molecules to form a more complex structure. In this case, boric acid and glycerin combine to form a complex \(\text{B(OH)}_3 \cdot \text{glycerin}\).
Complexes are often stabilized by interactions like hydrogen bonds or coordination bonds. The stability and formation of such complexes depend on factors like the concentrations of the components and their chemical compatibility.
  • Essential for processes like catalysis and enzyme-substrate interactions
  • Impacts properties like solubility and reactivity
  • Can be influenced by temperature, pressure, and concentration
Understanding how and why complexes form can help in controlling chemical reactions and designing chemical processes that rely on specific interactions.

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Most popular questions from this chapter

Pure \(\mathrm{PCl}_{5}\) gas is placed in a 2.00 -L. flask. After heating to \(250^{\circ} \mathrm{C}\) the pressure of \(\mathrm{PCl}_{5}\) is initially 2.000 atm. However, the gas slowly but only partially decomposes to gaseous \(\mathrm{PCl}_{3}\) and \(\mathrm{Cl}_{2}\). When equilibrium is reached, the partial pressure of \(\mathrm{Cl}_{2}\) is 0.814 atm. Calculate \(K_{\mathrm{p}}\) for the decomposition.

The equilibrium constant for the butane \(\rightleftarrows\) isobutane isomerization reaction is 2.5 at \(25^{\circ} \mathrm{C}\). If 1.75 mol of butane and 1.25 mol of isobutane are mixed, is the system at equilibrium? If not, when it proceeds to equilibrium, which reagent increases in concentration? Calculate the concentrations of the two compounds when the system reaches equilibrium.

The decomposition of \(\mathrm{NH}_{4} \mathrm{HS}\) $$ \mathrm{NH}_{4} \mathrm{HS}(\mathrm{s}) \rightleftharpoons \mathrm{NH}_{3}(\mathrm{g})+\mathrm{H}_{2} \mathrm{S}(\mathrm{g}) $$ is an endothermic process. Using Le Chatelier's principle, explain how increasing the temperature would affect the equilibrium. If more \(\mathrm{NH}_{4} \mathrm{HS}\) is added to a flask in which this equilibrium exists, how is the equilibrium affected? What if some additional \(\mathrm{NH}_{3}\) is placed in the flask? What will happen to the pressure of \(\mathrm{NH}_{3}\) if some \(\mathrm{H}_{2} \mathrm{S}\) is removed from the flask?

\(K_{\mathrm{p}}\) for the formation of phosgene, \(\mathrm{COCl}_{2},\) is \(6.5 \times 10^{11}\) at \(25^{\circ} \mathrm{C}\) $$ \mathrm{CO}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{g}) \rightleftharpoons \mathrm{COCl}_{2}(\mathrm{g}) $$ What is the value of \(K_{p}\) for the dissociation of phosgene? $$ \mathrm{COCl}_{2}(\mathrm{g}) \rightleftharpoons \mathrm{CO}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{g}) $$

Characterize each of the following as product- or reactant-favored at equilibrium. $$\begin{aligned} &\text { (a) } \mathrm{CO}(\mathrm{g})+1 / 2 \mathrm{O}_{2}(\mathrm{g}) \rightleftharpoons \mathrm{CO}_{2}(\mathrm{g})\\\ &&K_{\mathrm{p}}=1.2 \times 10^{45} \end{aligned}$$ $$\begin{aligned} &\text { (b) } \mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \rightleftharpoons \mathrm{H}_{2}(\mathrm{g})+1 / 2 \mathrm{O}_{2}(\mathrm{g})\\\ &&K_{\mathrm{p}}=9.1 \times 10^{-41} \end{aligned}$$ $$\begin{aligned} &\text { (c) } \mathrm{CO}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{g}) \rightleftharpoons \mathrm{COCl}_{2}(\mathrm{g})\\\ &&K_{\mathrm{p}}=6.5 \times 10^{11} \end{aligned}$$

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