Consider a weak acid, HX. If a 0.10-M solution of HX has a \(\mathrm{pH}\) 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
Calculating the acid dissociation constant (\(K_a\)) using the given pH and initial concentration of HX, we get \(K_a = \frac{(1.48 \times 10^{-6})^2}{(0.10 - 1.48 \times 10^{-6})}\). Next, we use the formula \(\Delta G^\circ = -RT \ln K_a\) to find the change in Gibbs free energy at 25°C. The result will indicate whether the reaction is spontaneous or not under standard conditions.

Step by step solution

01

Determine the acid dissociation constant (Ka)

First, use the given pH to determine the concentration of \(H^+\) ions. Remember, \(\text{pH} = -\log [H^+]\), so rearranging for \([H^+]\) gives \([H^+]=10^{-\text{pH}}\). Plugging in the given pH of 5.83, we find \([H^+] = 10^{-5.83}\) or \(1.48 \times 10^{-6}\) M. For a weak acid HX, it partially dissociates in water: \[HX <=> H^+ + X^-\]. At equilibrium, the concentration of \(H^+\) and \(X^-\) ions are equal, so \([X^-] = 1.48 \times 10^{-6}\) M. The dissociation constant of the acid, \(K_a\), is given by \(K_a = \frac{[H^+][X^-]}{[HX]}\). So using the given initial concentration of the acid \([HX]_{\text{initial}} = 0.10\) M and knowing that the \(HX\) that dissociates is equal to \([H^+]\), we get \([HX]_{\text{equilibrium}} = [HX]_{\text{initial}} - [H^+]\). Therefore, \(K_a = \frac{(1.48 \times 10^{-6})^2}{(0.10 - 1.48 \times 10^{-6})}\).
02

Calculate the standard Gibbs free energy change

After calculating \(K_a\), the next step is to use it to calculate the change in Gibbs free energy (\(\Delta G^\circ\)). The relationship between these two variables at standard conditions is given by the equation \(\Delta G^\circ = -RT \ln K_a\), where \(\Delta G^\circ\) is the standard Gibbs free energy, \(R\) is the ideal gas constant (equal to 8.314 J/(mol K) when \(G\) is expressed in Joules), and \(T\) is the temperature in Kelvin. First, convert the given temperature from Celsius to Kelvin by adding 273 to it (thus 25°C = 298 K). Then substitute these values into the equation: \(\Delta G^\circ = -8.314 \times 298 \times \ln K_a\).
03

Interpret the Result

The value obtained for \(\Delta G^\circ\) gives the change in Gibbs free energy under standard conditions for the dissociation reaction of the weak acid HX. If \(\Delta G^\circ\) is negative, it means the reaction is spontaneous under standard conditions. If \(\Delta G^\circ\) is positive, the reaction is non-spontaneous under standard conditions.

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Most popular questions from this chapter

Given the following data: $$2 \mathrm{H}_{2}(g)+\mathrm{C}(s) \longrightarrow \mathrm{CH}_{4}(g) \quad \Delta G^{\circ}=-51 \mathrm{kJ}$$ $$2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(l) \quad \Delta G^{\circ}=-474 \mathrm{kJ}$$ $$\mathrm{C}(s)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g) \quad \Delta G^{\circ}=-394 \mathrm{kJ}$$ Calculate \(\Delta G^{\circ}\) for \(\mathrm{CH}_{4}(g)+2 \mathrm{O}_{2}(g) \rightarrow \mathrm{CO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l)\)

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Consider the reactions $$\begin{array}{c} \mathrm{Ni}^{2+}(a q)+6 \mathrm{NH}_{3}(a q) \longrightarrow \mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6}^{2+}(a q) \\ \mathrm{Ni}^{2+}(a q)+3 \mathrm{en}(a q) \longrightarrow \mathrm{Ni}(\mathrm{en})_{3}^{2+}(a q) \end{array}$$ where $$\mathrm{en}=\mathrm{H}_{2} \mathrm{N}-\mathrm{CH}_{2}-\mathrm{CH}_{2}-\mathrm{NH}_{2}$$ The \(\Delta H\) values for the two reactions are quite similar, yet \(\mathrm{K}_{\text {reaction } 2}>K_{\text {reaction }} .\) Explain.

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Predict the sign of \(\Delta S^{\circ}\) for each of the following changes. a. \(\mathrm{K}(s)+\frac{1}{2} \mathrm{Br}_{2}(g) \longrightarrow \mathrm{KBr}(s)\) b. \(\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{NH}_{3}(g)\) c. \(\mathrm{KBr}(s) \longrightarrow \mathrm{K}^{+}(a q)+\mathrm{Br}^{-}(a q)\) d. \(\mathrm{KBr}(s) \longrightarrow \mathrm{KBr}(l)\)

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