Ch.16 - Aqueous Equilibria: Acids & Bases

Problem 158b

Chapter 16, Problem 158b

A 1.000 L sample of HF gas at 20.0 °C and 0.601 atm pressure was dissolved in enough water to make 50.0 mL of hydrofluoric acid. (b) To what volume must you dilute the solution to triple the percent dissociation?

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Determine the initial moles of HF gas using the ideal gas law: PV = nRT, where P is pressure, V is volume, n is moles, R is the ideal gas constant, and T is temperature in Kelvin.

Calculate the initial concentration of HF in the 50.0 mL solution by dividing the moles of HF by the volume of the solution in liters.

Understand that the percent dissociation of HF is related to its concentration and the equilibrium constant (Ka). Use the expression for percent dissociation: (dissociated concentration / initial concentration) * 100%.

To triple the percent dissociation, set up an equation where the new percent dissociation is three times the initial percent dissociation. Use the relationship between concentration and percent dissociation to express this condition.

Solve for the new volume of the solution that will achieve the desired percent dissociation by considering the dilution effect on concentration and using the equation C1V1 = C2V2, where C1 and V1 are the initial concentration and volume, and C2 and V2 are the final concentration and volume.

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Key Concepts

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

Percent Dissociation

Percent dissociation refers to the fraction of a weak acid or base that ionizes in solution, expressed as a percentage. For example, if 1 mole of HF dissociates into 0.3 moles of H+ and F-, the percent dissociation would be (0.3/1) * 100 = 30%. Understanding this concept is crucial for determining how dilution affects the ionization of the acid.

Dilution and Concentration

Dilution is the process of reducing the concentration of a solute in a solution, typically by adding more solvent. The relationship between the initial and final concentrations and volumes is given by the equation C1V1 = C2V2, where C is concentration and V is volume. This principle is essential for calculating the new volume required to achieve a desired percent dissociation.

Equilibrium Constant (Ka)

The equilibrium constant (Ka) for a weak acid quantifies its strength and the extent of dissociation in solution. It is defined as the ratio of the concentration of the products to the concentration of the reactants at equilibrium. For HF, the Ka value helps predict how changes in concentration, such as dilution, will affect the degree of ionization and thus the percent dissociation.

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Related Practice

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.

Textbook Question

A 1.000 L sample of HF gas at 20.0 °C and 0.601 atm pressure was dissolved in enough water to make 50.0 mL of hydrofluoric acid. (a) What is the pH of the solution?

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Open Question

A 200.0 mL sample of 0.350 M acetic acid (CH3CO2H) was allowed to react with 2.000 L of gaseous ammonia at 25 °C and a pressure of 650.8 mm Hg. Assuming no change in the volume of the solution, calculate the pH and the equilibrium concentrations of all species present (CH3CO2H, CH3CO2-, NH3, NH4+, H3O+, and OH-). Values of equilibrium constants are listed in Appendix C.

Textbook Question

You may have been told not to mix bleach and ammonia. The reason is that bleach (sodium hypochlorite) reacts with ammonia to produce toxic chloramines, such as NH2Cl. For example, in basic solution: OCl-1aq2 + NH31aq2S OH-1aq2 + NH2Cl1aq2 (b) The following mechanism has been proposed for this reaction in basic solution: H2O + OCl-HOCl + OH- Fast, equilibrium constantK1 HOCl + NH3 S H2O + NH2Cl Slow, rate constantk2 Assuming that the first step is in equilibrium and the second step is rate-determining, calculate the value of the rate constant k2 for the second step. Ka for HOCl is 3.5 * 10-8.

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