Problem 87 - An Introduction to Thermal Physics

Question

Sulfuric acid, \(\mathrm{H}_{2} \mathrm{SO}_{4},\) readily dissociates into \(\mathrm{H}^{+}\) and \(\mathrm{HSO}_{4}^{-}\) ions: $$\mathrm{H}_{2} \mathrm{SO}_{4} \longrightarrow \mathrm{H}^{+}+\mathrm{HSO}_{4}^{-}$$ The hydrogen sulfate ion, in turn, can dissociate again: $$\mathrm{HSO}_{4}^{-} \longmapsto \mathrm{H}^{+}+\mathrm{SO}_{4}^{2-}$$ 'The equilibrium constants for these reactions, in aqueous solutions at \(298 \mathrm{K},\) are approximately \(10^{2}\) and \(10^{-1.9}\), respectively. (For dissociation of acids it is usually more convenient to look up \(K\) than \(\Delta G^{\circ} .\) By the way, the negative base- 10 logarithm of \(K\) for such a reaction is called \(\mathbf{p K},\) in analogy to pH. So for the first reaction \(\mathrm{pK}=-2,\) while for the second reaction \(\mathrm{pK}=1.9 .2\) (a) Argue that the first reaction tends so strongly to the right that we might as well consider it to have gone to completion, in any solution that could possibly be considered dilute. At what pH values would a significant fraction of the sulfuric acid not be dissociated? (b) In industrialized regions where lots of coal is burned, the concentration of sulfate in rainwater is typically \(5 \times 10^{-5} \mathrm{mol} / \mathrm{kg}\). The sulfate can take any of the chemical forms mentioned above. Show that, at this concentration, the second reaction will also have gone essentially to completion, so all the sulfate is in the form of \(\mathrm{SO}_{4}^{2-} .\) What is the pH of this rainwater? (c) Explain why you can neglect dissociation of water into \(\mathrm{H}^{+}\) and \(\mathrm{OH}^{-}\) in answering the previous question. (d) At what pH would dissolved sulfate be equally distributed between HSO \(_{4}^{-}\) and \(\mathrm{SO}_{4}^{2-} ?\)

Step-by-Step Solution

Answer

(a) High pH does not cause incomplete dissociation of \(\mathrm{H}_2\mathrm{SO}_4\); (b) pH ≈ 4.3 for rainwater; (c) Water's \(\mathrm{H}^+\) dissociation is negligible; (d) Equal distribution at pH = 1.9.

Step 1: Analyze Reaction Equilibria

For part (a), the equilibrium constant for the first dissociation of sulfuric acid is given as 100 (or \(10^2\)), which is quite large. This indicates that the reaction strongly favors the product side, making it seem as though sulfuric acid dissociates completely in dilute solutions. A large \(K\) value suggests that almost all \(\mathrm{H}_2\mathrm{SO}_4\) molecules dissociate to form \(\mathrm{H}^+\) and \(\mathrm{HSO}_4^-\) ions.

Step 2: Consider Acidic and Neutral Conditions

Still addressing (a), due to the high equilibrium constant, the complete dissociation occurs at nearly all pH levels, except potentially very low pH where the high concentration of \(\mathrm{H}^+\) could suppress further dissociation. However, under typical conditions, any significant undissociated fraction is negligible. Hence, there aren't practically any pH levels where \(\mathrm{H}_2\mathrm{SO}_4\) remains undissociated.

Step 3: Evaluate the Second Dissociation at Low Concentration

For part (b), the concentration \(5 \times 10^{-5} \mathrm{mol/kg}\) is provided for \(\mathrm{SO}_4^{2-}\). Given that \(\mathrm{pK} = 1.9\) for the second dissociation step, we find \( K = 10^{-1.9} \approx 0.0126\), which typically suggests partial dissociation. However, at this concentration, the second reaction is driven by the low initial \(\mathrm{HSO}_4^-\) concentration leading to nearly complete conversion to \(\mathrm{SO}_4^{2-}\). Thus, at this low concentration,\(\mathrm{SO}_4^{2-}\) predominates.

Step 4: Calculate pH for Rainwater

To find the pH of rainwater in (b), note the completion of the second reaction indicates the dissolved hydrogen ions affecting the pH. The concentration of \(\mathrm{H}^+\) would be similar to the initial sulfate concentration, \(5 \times 10^{-5}\), yielding a pH of approximately 4.3 (since \(-\log(5 \times 10^{-5}) \approx 4.3\)).

Step 5: Justify Neglecting Water Dissociation

For part (c), the dissociation of water is negligible due to the significantly smaller contribution of \(\mathrm{H}^+\) from water compared to the acid dissociation. Since \( [\mathrm{H}^+] \gg 10^{-7} \) mol/L (the concentration from water), we ignore water's \(\mathrm{H}^+\) contribution.

Step 6: Determine Equal Distribution of Ion Species

For part (d), equal concentrations of \(\mathrm{HSO}_4^-\) and \(\mathrm{SO}_4^{2-}\) occur when the pH equals the \(\mathrm{pK}\) of the second dissociation step. Since \(\mathrm{pK} = 1.9\) for the second dissociation, \(\mathrm{pH} = 1.9\) results in equal amounts of \(\mathrm{HSO}_4^- \) and \(\mathrm{SO}_4^{2-}\).

Key Concepts

Acid Dissociation

In a chemical equilibrium, acid dissociation is a key concept, particularly for strong acids like sulfuric acid (\( \mathrm{H}_2\mathrm{SO}_4 \)).When sulfuric acid dissociates, it separates into \( \mathrm{H}^+ \) and \( \mathrm{HSO}_4^- \) ions. This initial dissociation is significant due to its complete nature at ordinary conditions, supported by a large equilibrium constant (\( K \approx 10^2 \)).A stronger (\( K \)) value implies a shift towards the right, favoring the formation of products.Following this, the hydrogen sulfate ion (\( \mathrm{HSO}_4^- \)) can further dissociate into \( \mathrm{H}^+ \) and sulfate ions (\( \mathrm{SO}_4^{2-} \)).This secondary dissociation is characteristically weaker, with an equilibrium constant of \( 10^{-1.9} \).Understanding acid dissociation helps in predicting what ions are present in a solution:

• Strong acids like sulfuric acid dissociate almost completely in dilute solutions.
• Weaker dissociation may still drive reactions towards completion, particularly under dilute conditions.

All these elements come into play when calculating how a solution's properties change with acid concentration and the respective equilibrium constants.

Equilibrium Constant

The equilibrium constant (\( K \)) plays a vital role in understanding the extent of a reaction at equilibrium. In the context of acid dissociation, \( K \) gives us insight into how completely a particular acid dissociates when dissolved in water.For sulfuric acid, the first dissociation step has a large \( K \approx 10^2 \), indicating almost complete dissociation. The reaction equilibrium strongly favors formation of \( \mathrm{H}^+ \) and \( \mathrm{HSO}_4^- \).

• A larger \( K \) value indicates a shift towards product formation.
• For strong acids, such significant shifts make certain reactions essentially go to completion.

For the second dissociation step (\( \mathrm{HSO}_4^- \rightarrow \mathrm{H}^+ + \mathrm{SO}_4^{2-} \)), the \( K \) is about \( 0.0126 \) (i.e., \( 10^{-1.9} \)).This implies a lesser but still notable tendency to dissociate under typical conditions.This concept becomes particularly relevant when considering the pH of solutions, determining whether reactions proceed completely, and is utilized in calculating important properties such as the pH.

pH Calculation

Calculating the pH of a solution involves determining the concentration of \( \mathrm{H}^+ \) ions, which influence the acidity of the solution. The pH is calculated using the formula:\[pH = -\log [\mathrm{H}^+]\]This formula suggests that knowing the equilibrium concentrations of hydrogen ions can guide the computation of pH directly.In our context, sulfuric acid dissociation contributes significantly to the \( \mathrm{H}^+ \) concentration:

• The initial complete dissociation of \( \mathrm{H}_2\mathrm{SO}_4 \) contributes highly to \( \mathrm{H}^+ \) concentration.
• The second dissociation step provides additional \( \mathrm{H}^+ \) but at a slower rate due to a smaller \( K \)

For example, given the typical concentration in industrial rainwater (\( 5 \times 10^{-5} \mathrm{mol/kg} \)), the rainwater's pH would be calculated to be around 4.3.Another essential aspect is realizing when certain dissociations can be considered complete — allowing simplifications in the pH calculation.Additionally, you must understand when the water dissociation is negligible compared to acid dissociations, as in most acidic solutions, \( [\mathrm{H}^+] \gg 10^{-7} \) mol/L.