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Chapter 17: Problem 125

Consider a weak acid, HX. If a \(0.10-M\) solution of \(\mathrm{HX}\) 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 can be calculated using the given pH value (5.83) and an equilibrium constant (Ka) for the dissociation reaction. By following the steps of finding the H⁺ concentration, OH⁻ concentration, calculating Ka, and then using the relationship between ΔG° and Ka, we find that ΔG° is approximately 67.9 kJ/mol.

Step by step solution

01

Write the dissociation reaction for the weak acid, HX

The dissociation reaction for a weak acid, HX, can be written as: \[ HX \rightleftharpoons H^+ + X^− \]
02

Calculate the concentration of hydronium ions (H⁺) from the given pH value

We are given that the pH of a 0.10 M solution of the weak acid is 5.83. We can use this information to find the concentration of H⁺ ions: \[ pH = - \log{[H^+]} \] Solving for [H⁺], we get: \[ [H^+] = 10^{−pH} = 10^{−5.83} \] \[ [H^+] \approx 1.49 \times 10^{-6} \,\mathrm{M} \]
03

Use ion-product constant of water (Kw) to find the hydroxide ions (OH⁻) concentration

We know the ion-product constant of water (Kw) at 25°C is: \[ K_{w} = [H^+][OH^−] = 1.00 \times 10^{-14} \] We can find the [OH⁻] concentration using the given [H⁺] concentration: \[ [OH^−] = \frac{K_{w}}{[H^+]} = \frac{1.00 \times 10^{-14}}{1.49 \times 10^{-6}} \] \[ [OH^−] \approx 6.71 \times 10^{-9}\, \mathrm{M} \]
04

Determine the dissociation constant, Ka, for the weak acid

We can write the acid dissociation constant (Ka) for the dissociation of HX as follows: \[ K_{a} = \frac{[H^+][X^−]}{[HX]} \] Since the acid is weak, we can assume that [X⁻] ≈ [H⁺]. The initial concentration of HX is 0.10 M, and the change in concentration due to dissociation is negligible. Hence, we can write: \[ K_{a} = \frac{(1.49 \times 10^{-6})^2}{0.10} \] \[ K_{a} \approx 2.22 \times 10^{-12} \]
05

Find the ΔG° for dissociation using the relation between ΔG° and Ka

We can find the ΔG° using the following relationship between the standard Gibbs free energy change (ΔG°) and the equilibrium constant (Ka): \[ ΔG° = -RT \ln{K_{a}} \] Where R is the gas constant (8.314 J/mol·K) and T is the temperature in Kelvin (298 K, since the temperature is 25°C). Plugging in the values, we obtain: \[ ΔG° = - (8.314)(298) \ln{(2.22 \times 10^{-12})} \] \[ ΔG° \approx 67\,924 \,\mathrm{J/mol} = 67.9\, \mathrm{kJ/mol} \] So, the standard Gibbs free energy change for the dissociation of the weak acid, HX, at 25°C is approximately 67.9 kJ/mol.

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Most popular questions from this chapter

Using Appendix 4 and the following data, determine \(S^{\circ}\) for $\mathrm{Fe}(\mathrm{CO})_{5}(g) . $\mathrm{Fe}(s)+5 \mathrm{CO}(g) \longrightarrow \mathrm{Fe}(\mathrm{CO})_{5}(g) \quad \Delta S^{\circ}=?$ $\mathrm{Fe}(\mathrm{CO})_{5}(l) \longrightarrow \mathrm{Fe}(\mathrm{CO})_{5}(g) \quad \Delta S^{\circ}=107 \mathrm{J} / \mathrm{K}$ $\mathrm{Fe}(s)+5 \mathrm{CO}(g) \longrightarrow \mathrm{Fe}(\mathrm{CO})_{5}(l) \quad \Delta S^{\circ}=-677 \mathrm{J} / \mathrm{K}$

Choose the substance with the larger positional probability in each case. a. 1 mole of \(\mathrm{H}_{2}\) (at \(\mathrm{STP} )\) or 1 mole of $\mathrm{H}_{2}\left(\text { at } 100^{\circ} \mathrm{C}, 0.5 \mathrm{atm}\right)$ b. 1 mole of \(\mathrm{N}_{2}(\text { at } \mathrm{STP})\) or 1 mole of \(\mathrm{N}_{2}(\text { at } 100 \mathrm{K}, 2.0 \mathrm{atm})\) c. 1 mole of \(\mathrm{H}_{2} \mathrm{O}(s)\) (at \(0^{\circ} \mathrm{C} )\) or 1 \(\mathrm{mole}\) of $\mathrm{H}_{2} \mathrm{O}(l)\left(\mathrm{at} 20^{\circ} \mathrm{C}\right)$

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Carbon tetrachloride \(\left(\mathrm{CCl}_{4}\right)\) and benzene \(\left(\mathrm{C}_{6} \mathrm{H}_{6}\right)\) form ideal solutions. Consider an equimolar solution of \(\mathrm{CC}_{4}\) and \(\mathrm{C}_{6} \mathrm{H}_{6}\) at \(25^{\circ} \mathrm{C}\) . The vapor above the solution is collected and condensed. Using the following data, determine the composition in mole fraction of the condensed vapor.

The enthalpy of vaporization of ethanol is 38.7 kJ/mol at its boiling point \(\left(78^{\circ} \mathrm{C}\right) .\) Determine $\Delta S_{\mathrm{sys}}, \Delta S_{\mathrm{surr}},\( and \)\Delta S_{\mathrm{univ}}$ when 1.00 mole of ethanol is vaporized at \(78^{\circ} \mathrm{C}\) and 1.00 atm.
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