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A \(0.250 \mathrm{M}\) solution of a weak base \(\mathrm{R}_{2} \mathrm{NH}\) has a \(\mathrm{pH}\) of 10.60 at \(25^{\circ} \mathrm{C}\). What is \(\Delta G^{\circ}\) for the dissociation of the weak base in water at \(25^{\circ} \mathrm{C}\) ? $$ \mathrm{R}_{2} \mathrm{NH}(a q)+\mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{R}_{2} \mathrm{NH}_{2}^{+}(a q)+\mathrm{OH}^{-}(a q) $$

Short Answer

Expert verified
Answer: The Gibbs free energy change (ΔG°) for the dissociation of the weak base R2NH in water at 25°C is 37.154 kJ/mol.

Step by step solution

01

Calculate pOH and hydroxide ion concentration

Given that the pH of the weak base solution is 10.60 at 25°C, we can find the pOH using the formula: pOH = 14 - pH pOH = 14 - 10.60 = 3.40 Now, we can find the hydroxide ion concentration [OH⁻] using the formula: [OH⁻] = 10^(-pOH) [OH⁻] = 10^(-3.40) = 3.98 x 10^(-4) M
02

Determine the value of Kb

In the reaction, the weak base R2NH accepts a proton from water to form R2NH2+ and OH⁻ ions. We are given the initial concentration of R2NH as 0.250 M. Let's denote the change in R2NH concentration as x. Thus, the equilibrium concentrations are: [R2NH] = 0.250 - x M [R2NH2+] = x M [OH⁻] = x M + 3.98 x 10^(-4) M Since we know [OH⁻] = 3.98 x 10^(-4) M at equilibrium, we can write the expression for Kb as: Kb = ([R2NH2+][OH⁻]) / [R2NH] Plugging in the equilibrium concentrations, we get: Kb = (x * (x + 3.98 * 10^(-4))) / (0.250 - x) x is small, so we can assume that x ≈ 0, which leads to: Kb ≈ (4.99 * 10^(-6)) / 0.250 Kb ≈ 1.998 x 10^(-5)
03

Calculate ΔG° using Kb

Now that we have obtained Kb, we can find ΔG° by using the following equation: ΔG° = -RT lnK where R is the gas constant (8.314 J/mol*K), T is the temperature (25°C = 298 K), and K is the equilibrium constant (Kb). ΔG° = -(8.314 J/mol*K) * (298 K) * ln(1.998 x 10^(-5)) ΔG° ≈ 37,154 J/mol Now, we can convert the ΔG° from J/mol to kJ/mol: ΔG° = 37.154 kJ/mol Thus, the Gibbs free energy change (ΔG°) for the dissociation of the weak base R2NH in water at 25°C is 37.154 kJ/mol.

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Key Concepts

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

pH and pOH Calculations
Understanding pH and pOH is crucial for students grappling with acid-base chemistry. Both of these values are integral in depicting the acidity or basicity of a solution. pH stands for 'potential of hydrogen' and is a measure of the hydrogen ion concentration in a solution. On the other hand, pOH represents the 'potential of hydroxide' and quantifies the hydroxide ion concentration.

In a neutral solution at 25°C, the pH is 7, which is considered neutral because it indicates an equal concentration of H⁺ and OH⁻ ions. For basic solutions, the pH will be above 7, and for acidic solutions, it will be below 7. The pH and pOH of a solution are related by the equation
\[ pH + pOH = 14 \]
This is known as the neutral point in the pH scale at room temperature. By using this relationship, if we know the pH, we can calculate the pOH, and vice versa. This helps us to understand the balance between hydrogen and hydroxide ions in solution which can then help predict the behavior of the solution in chemical reactions.
Hydroxide Ion Concentration
Delving deeper into the hydroxide ion (OH⁻) concentration, this part of the solution focuses on the basic characteristic of the weak base. Hydroxide ion concentration is a direct measure of the basicity of a solution. When dissolving a base in water, it reacts with water molecules and produces OH⁻ ions. The more OH⁻ ions present, the more basic the solution. For students, it is helpful to connect the pOH of the solution to the actual concentration of hydroxide ions. This is done through the equation:
\[ [OH⁻] = 10^{-pOH} \]
As pOH decreases, indicating a more basic solution, the power of ten becomes less negative, and the hydroxide ion concentration increases. It's like a see-saw: as one end goes up (the pH), the other goes down (the pOH), and vice versa. Working through the algebra, you can find the precise hydroxide ion concentration, which is pivotal in understanding the strength of a base in solution.
Gibbs Free Energy Change
Lastly, one of the most vital concepts in chemical thermodynamics is Gibbs free energy change (ΔG). It is the amount of energy capable of doing work during a chemical process when temperature and pressure are uniform. The equation for calculating ΔG° is:
\[ ΔG° = -RT \times \text{ln}K \]
where R is the universal gas constant, T is the temperature in Kelvin, and K is the equilibrium constant of the reaction, which can be either Kc, Ka, or Kb depending on the type of reaction. For a dissolution like the weak base, Kb is used.

Students should note that a negative ΔG° signifies a spontaneous process at constant temperature and pressure, while a positive value suggests a non-spontaneous process. In the context of our example, after calculating the Kb for our weak base, we can use it to find ΔG°, giving us insight into the energetic favorability of the base's dissociation in water.

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Most popular questions from this chapter

For the decomposition of \(\mathrm{Ag}_{2} \mathrm{O}\) : $$ 2 \mathrm{Ag}_{2} \mathrm{O}(s) \longrightarrow 4 \mathrm{Ag}(s)+\mathrm{O}_{2}(g) $$ (a) Obtain an expression for \(\Delta G^{\circ}\) as a function of temperature. Prepare a table of \(\Delta G^{\circ}\) values at \(100 \mathrm{~K}\) intervals between \(100 \mathrm{~K}\) and \(500 \mathrm{~K}\). (b) Calculate the temperature at which \(\Delta G^{\circ}=0\).

Which of the following processes are spontaneous? (a) cleaning up your desk (b) building a house of cards (c) sugar dissolving in hot water

In the laboratory, \(\mathrm{POCl}_{3}\) (phosphorus oxychloride) is used in the manufacture of phosphate esters, which are used in flame retardants and pesticides. It can be prepared by the following reaction at \(25^{\circ} \mathrm{C}\) : $$ \begin{array}{l} 2 \mathrm{PCl}_{3}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{POCl}_{3}(g) \\\ \Delta H^{\circ}=-572 \mathrm{~kJ} ; \Delta G^{\circ}=-518.7 \mathrm{~kJ} \end{array} $$ (a) Calculate \(\Delta S^{\circ} .\) Is the sign reasonable? (b) Calculate \(S^{\circ}\) for \(\mathrm{POCl}_{3}\). (c) Calculate \(\Delta H_{\mathrm{f}}^{\circ}\) for \(\mathrm{POCl}_{3}\).

Consider the following reactions at \(25^{\circ} \mathrm{C}\) : $$ \begin{array}{r} \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(a q)+6 \mathrm{O}_{2}(g) \longrightarrow 6 \mathrm{CO}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O} \\ \Delta G^{\circ}=-2870 \mathrm{~kJ} \\ \mathrm{ADP}(a q)+\mathrm{HPO}_{4}^{2-}(a q)+2 \mathrm{H}^{+}(a q) \longrightarrow \mathrm{ATP}(a q)+\mathrm{H}_{2} \mathrm{O} \\ \Delta G^{\circ}=31 \mathrm{~kJ} \end{array} $$ Write an equation for a coupled reaction between glucose, \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6},\) and \(\mathrm{ADP}\) in which \(\Delta G^{\circ}=-390 \mathrm{~kJ}\).

Given the following standard free energies at \(25^{\circ} \mathrm{C}\) for the following reactions: $$ \begin{aligned} \mathrm{N}_{2} \mathrm{O}_{5}(g) \longrightarrow & 2 \mathrm{NO}(g)+\frac{3}{2} \mathrm{O}_{2}(g) & & \Delta G^{\circ}=-59.2 \mathrm{~kJ} \\ \mathrm{NO}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{NO}_{2}(g) & & \Delta G^{\circ}=-35.6 \mathrm{~kJ} \end{aligned} $$ Calculate \(\Delta G^{\circ}\) at \(25^{\circ} \mathrm{C}\) for the following reaction: $$ 2 \mathrm{NO}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{N}_{2} \mathrm{O}_{5}(g) $$

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