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Chapter 17: Problem 107

Consider a weak acid, HX. If a \(0.10 M\) solution of HX has a pH of \(5.83\) at \(25^{\circ} \mathrm{C}\), what is \(\Delta G^{\circ}\) for the acid's dissociation reaction at \(25^{\circ} \mathrm{C}\) ?

Short Answer

Expert verified
The standard Gibbs Free Energy change (ΔG°) for the dissociation of the weak acid HX at 25°C is 56.196 kJ/mol.

Step by step solution

01

Convert the pH to the concentration of H+ ions

We first need to find the concentration of H+ ions from the given pH value: pH = - log[H+] Or [H+] = 10^(-pH) Given pH = 5.83, we can determine the concentration of H+ ions: [H+] = 10^(-5.83) = 1.47 × 10^(-6) M
02

Write the chemical equation for the dissociation of HX

The chemical equation for the dissociation of the weak acid HX is: HX(aq) ⇌ H+(aq) + X^-(aq)
03

Calculate the equilibrium constant (Ka) for the reaction

Using the reaction above, we can now determine the equilibrium constant (Ka). Since the dissociation is at equilibrium, we can write the expression for Ka: \[ K_a = \frac{[H+][X^-]}{[HX]} \] We know that [H+] = 1.47 × 10^(-6) M, and [HX] at the start is 0.10 M. Since the weak acid dissociates partially, at equilibrium, the concentration of X⁻ will also be equal to [H+]. Therefore, [X⁻] = 1.47 × 10^(-6) M. Now, we can calculate Ka: \[ K_a = \frac{(1.47 \times 10^{-6})(1.47 \times 10^{-6})}{0.10} \] Ka = 2.15 × 10^(-12)
04

Determine the standard reaction Gibbs energy change (ΔG°) using the relationship between ΔG° and Ka

Finally, we need to find the standard Gibbs Free Energy change (ΔG°) using the following formula: \[ \Delta G^{\circ} = -RT \ln(K_a) \] Where R is the gas constant (8.314 J/(mol·K)) and T is the temperature in Kelvin (25°C = 298.15 K) \[ \Delta G^{\circ} = -(8.314)(298.15) \ln(2.15 \times 10^{-12}) \] ΔG° = 56,196 J/mol = 56.196 kJ/mol Therefore, the standard Gibbs Free Energy change (ΔG°) for the dissociation of the weak acid HX at 25°C is 56.196 kJ/mol.

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Most popular questions from this chapter

What types of experiments can be carried out to determine whether a reaction is spontaneous? Does spontaneity have any relationship to the final equilibrium position of a reaction? Explain.

Many biochemical reactions that occur in cells require relatively high concentrations of potassium ion \(\left(\mathrm{K}^{+}\right)\). The concentration of \(\mathrm{K}^{+}\) in muscle cells is about \(0.15 M\). The concentration of \(\mathrm{K}^{+}\) in blood plasma is about \(0.0050 M .\) The high internal concentration in cells is maintained by pumping \(\mathrm{K}^{+}\) from the plasma. How much work must be done to transport \(1.0 \mathrm{~mol} \mathrm{~K}^{+}\) from the blood to the inside of a muscle cell at \(37^{\circ} \mathrm{C}\), normal body temperature? When \(1.0 \mathrm{~mol} \mathrm{~K}^{+}\) is transferred from blood to the cells, do any other ions have to be transported? Why or why not?

The synthesis of glucose directly from \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\) and the synthesis of proteins directly from amino acids are both nonspontaneous processes under standard conditions. Yet it is necessary for these to occur for life to exist. In light of the second law of thermodynamics, how can life exist?

Two crystalline forms of white phosphorus are known. Both forms contain \(\mathrm{P}_{4}\) molecules, but the molecules are packed together in different ways. The \(\alpha\) form is always obtained when the liquid freezes. However, below \(-76.9^{\circ} \mathrm{C}\), the \(\alpha\) form spontaneously converts to the \(\beta\) form: $$\mathrm{P}_{4}(s, \alpha) \longrightarrow \mathrm{P}_{4}(s, \beta)$$ a. Predict the signs of \(\Delta H\) and \(\Delta S\) for this process. b. Predict which form of phosphorus has the more ordered crystalline structure (has the smaller positional probability).

Predict the sign of \(\Delta S_{\text {surr }}\) for the following processes. a. \(\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{H}_{2} \mathrm{O}(g)\) b. \(I_{2}(g) \longrightarrow I_{2}(s)\)
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