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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 (螖G掳) for the dissociation of the weak acid HX at 25掳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鈦� concentration, OH鈦� concentration, calculating Ka, and then using the relationship between 螖G掳 and Ka, we find that 螖G掳 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: \[ HX \rightleftharpoons H^+ + X^鈭� \]
02

Calculate the concentration of hydronium ions (H鈦�) 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鈦� ions: \[ pH = - \log{[H^+]} \] Solving for [H鈦篯, we get: \[ [H^+] = 10^{鈭抪H} = 10^{鈭�5.83} \] \[ [H^+] \approx 1.49 \times 10^{-6} \,\mathrm{M} \]
03

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

We know the ion-product constant of water (Kw) at 25掳C is: \[ K_{w} = [H^+][OH^鈭抅 = 1.00 \times 10^{-14} \] We can find the [OH鈦籡 concentration using the given [H鈦篯 concentration: \[ [OH^鈭抅 = \frac{K_{w}}{[H^+]} = \frac{1.00 \times 10^{-14}}{1.49 \times 10^{-6}} \] \[ [OH^鈭抅 \approx 6.71 \times 10^{-9}\, \mathrm{M} \]
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: \[ K_{a} = \frac{[H^+][X^鈭抅}{[HX]} \] Since the acid is weak, we can assume that [X鈦籡 鈮� [H鈦篯. 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 螖G掳 for dissociation using the relation between 螖G掳 and Ka

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

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Key Concepts

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

Weak Acid Dissociation
When discussing weak acid dissociation, we're looking at how a weak acid breaks down into its ions in a solution. Unlike strong acids, which dissociate completely in solution, weak acids only partially dissociate. This means that at equilibrium, you have both the undissociated acid and its ions present.

Take for instance a weak acid denoted as HX. In water, it dissociates into hydrogen ions ( H^+ ) and the conjugate base ( X^- ). This can be represented by the reversible reaction:
  • HX 鈬� H^+ + X^鈭�
This equilibrium is dynamic, meaning the forward and reverse reactions occur simultaneously but at the same rate, ensuring the concentrations of reactants and products remain constant. Because weak acids do not fully dissociate, the concentration of H^+ ions is smaller compared to the initial concentration of the acid. This partial dissociation is key to understanding the behavior and calculations related to weak acids, especially when assessing their strength and effect in a solution.
Acid Dissociation Constant
The acid dissociation constant, denoted as K_a, is an essential concept for quantifying the strength of a weak acid. It is a measure of how well an acid dissociates into its ions in a solution. The larger the K_a value, the stronger the acid, because it means the acid dissociates more completely.
  • The formula for K_a is given by:\[ K_{a} = \frac{[H^+][X^-]}{[HX]} \]This equation shows that K_a is the ratio of the concentrations of the products (hydrogen ions and conjugate base) to the undissociated acid.
  • Since we often deal with very small numbers when talking about weak acids, K_a values are typically converted to pK_a (the negative logarithm of K_a) for ease of use: \[ pK_a = -\log{K_a} \]
Calculating K_a helps us understand and predict how an acid will behave in diverse situations, which is crucial in many scientific and industrial applications. For the weak acid HX discussed before, by knowing the concentration of H^+ in solution, one can easily compute K_a to appreciate how much the acid has dissociated.
pH and Hydronium Ion Concentration
Understanding the relationship between pH and hydronium ion concentration is fundamental in chemistry. The pH scale, ranging from 0 to 14, is used to measure the acidity or basicity of a solution. It is directly linked to the concentration of hydronium ions ([H^+]).
  • The formula to calculate pH is:\[ pH = - \log{[H^+]} \]This translates small concentrations of hydrogen ions into a simple figure between 0 and 14. At pH = 7, the solution is neutral; if pH < 7, it鈥檚 acidic; and if pH > 7, it鈥檚 basic.
  • Given a pH value, you can also reverse the process to find the hydronium ion concentration by using:\[ [H^+] = 10^{-pH} \]
For the weak acid HX example given, a pH of 5.83 was used to determine that the [H^+] concentration is approximately 1.49 \times 10^{-6} \, \text{M}, illustrating how even a weak acid can influence the pH of a solution significantly. Knowing the pH allows for better control and adjustment in experiments and industrial processes, making it a key parameter in chemical reactions.

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Most popular questions from this chapter

For each of the following pairs, which substance has the greater value of S? a. \(\mathrm{N}_{2} \mathrm{O}(\text { at } 0 \mathrm{K})\) or He (at 10 \(\mathrm{K} )\) b. \(\mathrm{N}_{2} \mathrm{O}(g)\) (at \(1 \mathrm{atm}, 25^{\circ} \mathrm{C} )\) or He(g) (at 1 atm, \(25^{\circ} \mathrm{C} )\) c. \(\mathrm{NH}_{3}(s)\) (at 196 \(\mathrm{K} ) \longrightarrow \mathrm{NH}_{3}(l)(\text { at } 196 \mathrm{K})\)

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.

Human DNA contains almost twice as much information as is needed to code for all the substances produced in the body. Likewise, the digital data sent from Voyager II contained one redundant bit out of every two bits of information. The Hubble space telescope transmits three redundant bits for every bit of information. How is entropy related to the transmission of information? What do you think is accomplished by having so many redundant bits of information in both DNA and the space probes?

The equilibrium constant for a certain reaction decreases from 8.84 to \(3.25 \times 10^{-2}\) when the temperature increases from \(25^{\circ} \mathrm{C}\) to \(75^{\circ} \mathrm{C}\) . Estimate the temperature where \(K=1.00\) for this reaction. Estimate the value of \(\Delta S^{\circ}\) for this reaction. (Hint: Manipulate the equation in Exercise 85.)

The third law of thermodynamics states that the entropy of a perfect crystal at 0 \(\mathrm{K}\) is zero. In Appendix \(4, \mathrm{F}^{-}(a q), \mathrm{OH}^{-}(a q)\) and \(\mathrm{S}^{2-}(a q)\) all have negative standard entropy values. How can \(S^{\circ}\) values be less than zero?
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