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
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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 (loneprotons, H+ (aq), do not exist!).
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(a) Write the equilibrium constant expression for the dissociation of acetic acid.
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(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?
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(c) What is the value of the equilibrium constant Keq for the dissociation according to the concentrations from part (b)?
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(d) When starting with completely un-dissociated acetic acid, is it accurate to assume that [HOAc]0 = [HOAc]eq? Why or why not?
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(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?
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(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?
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