Soft drinks such as Coca Cola and Pepsi are acidic due to the presence of dissolved carbon dioxide that forms carbonic acid. Assume none of the carbonic acid exists as dissolved carbon dioxide. The dissociation of carbonic acid is: H;COs(aq) =HCO3 (aq) + H*(aq) For the problem, assume that the bicarbonate ion, HCO3', does not dissociate again to form the carbonate ion. In order to determine the acidity of a 355 mL can of soda, 0.500 M NaOH is titrated in, taking 25.78 mL to fully neutralize the carbonic acid. This leads to the result that there was 0.0129 mol H2CO3 in the can, with an initial pH of 2.32. (a) After the titration, what major species are present in the solution, and which will dominate the pH? (b) Which of the following acid-base equilibrium reactions determines the pH after the titration? I. H2CO3(aq) =HCO3 (aq) + H*(aq) Ka = 2.51×104 %3D II. HCO3'(aq) + Hz0(1) = H2CO3(aq) + OH'(aq) %3D (c) What is the pH of the solution after the titration?
Soft drinks such as Coca Cola and Pepsi are acidic due to the presence of dissolved carbon dioxide that forms carbonic acid. Assume none of the carbonic acid exists as dissolved carbon dioxide. The dissociation of carbonic acid is: H;COs(aq) =HCO3 (aq) + H*(aq) For the problem, assume that the bicarbonate ion, HCO3', does not dissociate again to form the carbonate ion. In order to determine the acidity of a 355 mL can of soda, 0.500 M NaOH is titrated in, taking 25.78 mL to fully neutralize the carbonic acid. This leads to the result that there was 0.0129 mol H2CO3 in the can, with an initial pH of 2.32. (a) After the titration, what major species are present in the solution, and which will dominate the pH? (b) Which of the following acid-base equilibrium reactions determines the pH after the titration? I. H2CO3(aq) =HCO3 (aq) + H*(aq) Ka = 2.51×104 %3D II. HCO3'(aq) + Hz0(1) = H2CO3(aq) + OH'(aq) %3D (c) What is the pH of the solution after the titration?
Chemistry: Principles and Practice
3rd Edition
ISBN:9780534420123
Author:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward Mercer
Publisher:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward Mercer
Chapter18: Electrochemistry
Section: Chapter Questions
Problem 18.103QE
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Since you have posted question with multiple sub-parts, we are entitled to answer the first 3 only.
Given: Moles of H2CO3 in the can = 0.0129 mol.
Volume of soda can used = 355 mL = 0.355 L
Volume of NaOH solution used = 25.78 mL = 0.02578 L
And HCO3- does not dissociate further to produce CO32-.
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