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Ch.16 - Aqueous Equilibria: Acids & Bases
Chapter 16, Problem 153

In the case of very weak acids, 3H3O+ 4 from the dissociation of water is significant compared with 3H3O+ 4 from the dissociation of the weak acid. The sugar substitute saccharin 1C7H5NO3S2, for example, is a very weak acid having Ka = 2.1 * 10-12 and a solubility in water of 348 mg/100 mL. Calculate 3H3O+ 4 in a saturated solution of saccharin. (Hint: Equilibrium equations for the dissociation of saccharin and water must be solved simultaneously.)

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1
Convert the solubility of saccharin from mg/100 mL to molarity (mol/L) by using its molar mass.
Write the equilibrium expression for the dissociation of saccharin in water: \( \text{C}_7\text{H}_5\text{NO}_3\text{S}_2 \rightleftharpoons \text{H}^+ + \text{C}_7\text{H}_4\text{NO}_3\text{S}_2^- \).
Write the equilibrium expression for the dissociation of water: \( \text{H}_2\text{O} \rightleftharpoons \text{H}^+ + \text{OH}^- \).
Set up the equilibrium expressions for both reactions: \( K_a = \frac{[\text{H}^+][\text{C}_7\text{H}_4\text{NO}_3\text{S}_2^-]}{[\text{C}_7\text{H}_5\text{NO}_3\text{S}_2]} \) and \( K_w = [\text{H}^+][\text{OH}^-] \).
Solve the system of equations simultaneously to find the concentration of \([\text{H}^+]\) in the saturated solution.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Weak Acids and Their Dissociation

Weak acids are substances that partially dissociate in solution, establishing an equilibrium between the undissociated acid and its ions. The strength of a weak acid is quantified by its acid dissociation constant (Ka), which indicates the extent of dissociation. For saccharin, with a Ka of 2.1 x 10^-12, this means that only a small fraction of the acid molecules donate protons (H+) to the solution, making the calculation of hydronium ion concentration (H3O+) crucial.
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Equilibrium and the ICE Table

In chemical equilibria, the concentrations of reactants and products remain constant over time. The ICE (Initial, Change, Equilibrium) table is a useful tool for organizing the concentrations of species involved in a reaction at different stages. For the dissociation of saccharin and water, setting up an ICE table allows for the systematic calculation of H3O+ concentrations by accounting for the contributions from both the weak acid and the autoionization of water.
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Simultaneous Equations in Equilibrium Calculations

When dealing with multiple equilibria, such as the dissociation of a weak acid and the autoionization of water, simultaneous equations are often necessary to find the concentrations of all species involved. In this case, the concentration of H3O+ from both the dissociation of saccharin and the water must be considered together. Solving these equations allows for a comprehensive understanding of the system's behavior and the accurate calculation of the hydronium ion concentration in the saturated solution.
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Related Practice
Open Question
Sulfur dioxide is quite soluble in water: SO2(g) + H2O(l) ⇌ H2SO3(aq), K = 1.33. The H2SO3 produced is a weak diprotic acid (Ka1 = 1.5 * 10^-2; Ka2 = 6.3 * 10^-8). Calculate the pH and the concentrations of H2SO3, HSO3-, and SO3^2- in a solution prepared by continuously bubbling SO2 at a pressure of 1.00 atm into pure water.
Textbook Question
Acid and base behavior can be observed in solvents other than water. One commonly used solvent is dimethyl sulfoxide (DMSO), which can be treated as a monoprotic acid 'HSol.' Just as water can behave either as an acid or a base, so HSol can behave either as a Brønsted–Lowry acid or base. (b) The weak acid HCN has an acid dissociation constant Ka = 1.3 * 10-13 in the solvent HSol. If 0.010 mol of NaCN is dissolved in 1.00 L of HSol, what is the equilibrium concentration of H2Sol + ?
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Textbook Question
A 7.0 mass % solution of H3PO4 in water has a density of 1.0353 g/mL. Calculate the pH and the molar concentrations of all species present (H3PO4, H2PO4-, PO43-, H3O+ , and OH-) in the solution. Values of equilibrium constants are listed in Appendix C.
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Textbook Question
In aqueous solution, sodium acetate behaves as a strong electrolyte, yielding Na+ cations and CH3CO2 - anions. A particular solution of sodium acetate has a pH of 9.07 and a density of 1.0085 g/mL. What is the molality of this solution, and what is its freezing point?
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Textbook Question

During a certain time period, 4.0 million tons of SO2 were released into the atmosphere and subsequently oxidized to SO3. As explained in the Inquiry, the acid rain produced when the SO3 dissolves in water can damage marble statues: CaCO3(s) + H2SO4(aq) → CaSO4(aq) + CO2(g) + H2O(l) (a) How many 500 pound marble statues could be damaged by the acid rain? (Assume that the statues are pure CaCO3 and that a statue is damaged when 3.0% of its mass is dissolved.)

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Open Question
We’ve said that alkali metal cations do not react appreciably with water to produce H3O+ ions, but in fact, all cations are acidic to some extent. The most acidic alkali metal cation is the smallest one, Li+, which has Ka = 2.5 * 10^-14 for the reaction: Li(H2O)4+ (aq) + H2O (l) ⇌ H3O+ (aq) + Li(H2O)3(OH) (aq). This reaction and the dissociation of water must be considered simultaneously in calculating the pH of Li+ solutions, which nevertheless have pH ≈ 7. Check this by calculating the pH of a 0.10 M LiCl solution.