Textbook Question
An acid solution with a concentration of 0.500 M has a
pH = 3.21. What is the Ka of the acid? (LO 16.8)
(a) 1.2 * 10-5 (b) 1.7 * 10-6
(c) 7.6 * 10-7 (d) 5.4 * 10-3
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Ch.16 - Aqueous Equilibria: Acids & Bases
Chapter 16, Problem 12
Using values of Kb in Appendix C, calculate values of Ka for each of the following ions: (b) Hydroxylammonium ion, NH3OH+; (c) Anilinium ion, C6H5NH3+; (d) Pyridinium ion, C5H5NH+.
Verified step by step guidance1
Step 1: Understand the relationship between Ka and Kb. For a conjugate acid-base pair, the product of the acid dissociation constant (Ka) and the base dissociation constant (Kb) is equal to the ion product of water (Kw), which is 1.0 x 10^-14 at 25°C. This can be expressed as: Ka * Kb = Kw.
Step 2: Identify the conjugate base for each ion. For hydroxylammonium ion (NH3OH+), the conjugate base is hydroxylamine (NH2OH). For anilinium ion (C6H5NH3+), the conjugate base is aniline (C6H5NH2). For pyridinium ion (C5H5NH+), the conjugate base is pyridine (C5H5N).
Step 3: Look up the Kb values for the conjugate bases in Appendix C. These values are necessary to calculate the Ka values for the corresponding conjugate acids.
Step 4: Calculate the Ka for each ion using the formula: Ka = Kw / Kb. Substitute the known values of Kw and the Kb for each conjugate base into the formula to find the Ka for each ion.
Step 5: Verify the units and ensure that the calculated Ka values are consistent with the expected order of magnitude for weak acids, typically ranging from 10^-1 to 10^-14.
Related Practice
Open Question
Determine the following concentrations for a 0.40 M H2Se solution that has the stepwise dissociation constants of Ka1 = 1.3 * 10^-4 and Ka2 = 1.0 * 10^-11. (LO 16.10) (a) [H2Se] = 0.35, [HSe^-] = 5.0 * 10^-2, [H3O^+] = 3.0 * 10^-3, [Se^2-] = 1.3 * 10^-4 (b) [H2Se] = 0.39, [HSe^-] = 7.2 * 10^-3, [H3O^+] = 7.2 * 10^-3, [Se^2-] = 1.0 * 10^-11 (c) [H2Se] = 0.31, [HSe^-] = 9.0 * 10^-2, [H3O^+] = 9.0 * 10^-2, [Se^2-] = 1.0 * 10^-11 (d) [H2Se] = 0.40, [HSe^-] = 1.3 * 10^-4, [H3O^+] = 1.3 * 10^-4, [Se^2-] = 1.0 * 10^-11
Textbook Question
Ammonia 1NH32 has base dissociation constant 1Kb2 of
1.8 * 10-5. What is the concentration of an aqueous ammonia
solution that has a pH of 11.68? (LO 16.11)
(a) 0.28 M (b) 3.6 M
(c) 9.0 * 10-3 M (d) 1.3 M
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Textbook Question
Consider the reaction: SO2 + OH- S HSO3
-. Which reaction scheme shows the correct use of the curved arrow notation representing the
donation of an electron pair and the correct labeling of the Lewis acid and Lewis base? (LO 16.14)
(a)
(b)
(c)
(d)
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Open Question
For each of the Lewis acid–base reactions in Problem 16.138, draw electron-dot structures for the reactants and products, and use the curved arrow notation to represent the donation of a lone pair of electrons from the Lewis base to the Lewis acid.
Open Question
For a solution of two weak acids with comparable values of Ka, there is no single principal reaction. The two acid dissociation equilibrium equations must therefore be solved simultaneously. Calculate the pH in a solution that is 0.10 M in acetic acid (CH3CO2H, Ka = 1.8 * 10^-5) and 0.10 M in benzoic acid (C6H5CO2H, Ka = 6.5 * 10^-5). (Hint: Let x = [CH3CO2H] that dissociates and y = [C6H5CO2H] that dissociates; then [H3O+] = x + y.)