EBK INTRODUCTORY CHEMISTRY
8th Edition
ISBN: 9780134553153
Author: CORWIN
Publisher: PEARSON CO
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Chapter 16, Problem 9CE
Interpretation Introduction
Interpretation:
The equilibrium constant for slightly soluble ionic compound,
Concept introduction:
Solubility product constant,
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Acetic acid is a weak acid, meaning it does not fully dissociate in water. Instead, there is an equilibrium between the dissolved but undissociated molecule and the component ions:
HOAc (aq) + H2O (l) ⇌ H3O+ (aq) + OAc– (aq)OAc– is an abbreviation for the acetate ion, CH3COO–, and H3O+ is the hydronium ion (lone protons, H+ (aq), do not exist!).
(d) When starting with completely un-dissociated acetic acid, is it accurate to assume that [HOAc]0 = [HOAc]eq? Why or why not?
(e) A highly concentrated acetic acid solution contains 15.0M acetic acid at equilibrium. What are the equilibrium concentrations of the hydronium and acetate ions in this solution?
(f) Creating the concentrated acetic acid solution by dissolving liquid HOAc in water raises the temperature of the water by about 5°C from room temperature. At 50°C, do you expect the solution to contain more or less acetate ion OAc– than what you calculated in (c)? Why?
Acetic acid is a weak acid, meaning it does not fully dissociate in water. Instead, there is an equilibrium between the dissolved but undissociated molecule and the component ions:
HOAc (aq) + H2O (l) ⇌ H3O+ (aq) + OAc– (aq)OAc– is an abbreviation for the acetate ion, CH3COO–, and H3O+ is the hydronium ion (lone
protons, H+ (aq), do not exist!).
(a) Write the equilibrium constant expression for the dissociation of acetic acid.
(b) Vinegar sold commercially is typically 0.8 − 1.0 M acetic acid. A 1.00 M solution of acetic acid is measured by its pH to have an equilibrium concentration of 4.19×10−3 M for both acetate ions and hydronium ions at room temperature. Assuming [HOAc]0 = 1.00M, what is the equilibrium concentration of undissociated acetic acid [HOAc]eq to the correct number of significant figures?
(c) What is the value of the equilibrium constant Keq for the dissociation according to the concentrations from part (b)?
(d) When starting with completely un-dissociated…
Write the equilibrium constant expression for this reaction:
NH(aq) → NH3(aq) + H+ (aq)
-
x
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Chapter 16 Solutions
EBK INTRODUCTORY CHEMISTRY
Ch. 16 - Prob. 1CECh. 16 - Prob. 2CECh. 16 - Prob. 3CECh. 16 - Prob. 4CECh. 16 - Prob. 5CECh. 16 - Prob. 6CECh. 16 - Prob. 7CECh. 16 - Prob. 8CECh. 16 - Prob. 9CECh. 16 - Prob. 10CE
Ch. 16 - Prob. 1KTCh. 16 - Prob. 2KTCh. 16 - Prob. 3KTCh. 16 - Prob. 4KTCh. 16 - Prob. 5KTCh. 16 - Prob. 6KTCh. 16 - Prob. 7KTCh. 16 - Prob. 8KTCh. 16 - Prob. 9KTCh. 16 - Prob. 10KTCh. 16 - Prob. 11KTCh. 16 - Prob. 12KTCh. 16 - Prob. 13KTCh. 16 - Prob. 14KTCh. 16 - Prob. 15KTCh. 16 - Prob. 16KTCh. 16 - Prob. 17KTCh. 16 - Prob. 18KTCh. 16 - Prob. 1ECh. 16 - Prob. 2ECh. 16 - Prob. 3ECh. 16 - Prob. 4ECh. 16 - Prob. 5ECh. 16 - Prob. 6ECh. 16 - Prob. 7ECh. 16 - Prob. 8ECh. 16 - Prob. 9ECh. 16 - Prob. 10ECh. 16 - Prob. 11ECh. 16 - Prob. 12ECh. 16 - Prob. 13ECh. 16 - Prob. 14ECh. 16 - Prob. 15ECh. 16 - Prob. 16ECh. 16 - Prob. 17ECh. 16 - Prob. 18ECh. 16 - Prob. 19ECh. 16 - Prob. 20ECh. 16 - Prob. 21ECh. 16 - Prob. 22ECh. 16 - Prob. 23ECh. 16 - Prob. 24ECh. 16 - Prob. 25ECh. 16 - Prob. 26ECh. 16 - Prob. 27ECh. 16 - Prob. 28ECh. 16 - Prob. 29ECh. 16 - Prob. 30ECh. 16 - Prob. 31ECh. 16 - Prob. 32ECh. 16 - Prob. 33ECh. 16 - Prob. 34ECh. 16 - Prob. 35ECh. 16 - Prob. 36ECh. 16 - Prob. 37ECh. 16 - Prob. 38ECh. 16 - Prob. 39ECh. 16 - Prob. 40ECh. 16 - Prob. 41ECh. 16 - Prob. 42ECh. 16 - Prob. 43ECh. 16 - Prob. 44ECh. 16 - Prob. 45ECh. 16 - Prob. 46ECh. 16 - Prob. 47ECh. 16 - Prob. 48ECh. 16 - Prob. 1STCh. 16 - Prob. 2STCh. 16 - Prob. 3STCh. 16 - Prob. 4STCh. 16 - Prob. 5STCh. 16 - Prob. 6STCh. 16 - Prob. 7STCh. 16 - Prob. 8STCh. 16 - Prob. 9STCh. 16 - Prob. 10STCh. 16 - Prob. 11STCh. 16 - Prob. 12STCh. 16 - Prob. 13STCh. 16 - Prob. 14STCh. 16 - Prob. 15ST
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