“A typical vitamin C tablet contains H2C6H6O6. If two tablets are dissolved in about a cup of water, the resulting concentration of the ascorbic acid is 0.0115 M.” i gave the equation for the first ionization dissociation, and was given the Ka of 8.00x10^-5. At first I assumed .0115 was the equilibrium concentration so I solved that way, but wound up with an x that would give me a negative concentration for the second ionization, so then I solved using 0.0115 as the initial concentration, and had the same issue, getting x=9.20x10^-4. For the second ionization I’d be taking that and then subtracting the new x, which I solved as 3.9X10^-3, which again gives me a negative concentration. Please help!
“A typical vitamin C tablet contains H2C6H6O6. If two tablets are dissolved in about a cup of water, the resulting concentration of the ascorbic acid is 0.0115 M.” i gave the equation for the first ionization dissociation, and was given the Ka of 8.00x10^-5. At first I assumed .0115 was the equilibrium concentration so I solved that way, but wound up with an x that would give me a negative concentration for the second ionization, so then I solved using 0.0115 as the initial concentration, and had the same issue, getting x=9.20x10^-4. For the second ionization I’d be taking that and then subtracting the new x, which I solved as 3.9X10^-3, which again gives me a negative concentration. Please help!
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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“A typical vitamin C tablet contains H2C6H6O6. If two tablets are dissolved in about a cup of water, the resulting concentration of the ascorbic acid is 0.0115 M.”
i gave the equation for the first ionization dissociation, and was given the Ka of 8.00x10^-5.
At first I assumed .0115 was the equilibrium concentration so I solved that way, but wound up with an x that would give me a negative concentration for the second ionization, so then I solved using 0.0115 as the initial concentration, and had the same issue, getting x=9.20x10^-4. For the second ionization I’d be taking that and then subtracting the new x, which I solved as 3.9X10^-3, which again gives me a negative concentration. Please help!
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