0.030 moles of a weak acid, HA, was dissolved in 2.0 L of water to form a solution. At equilibrium, the concentration of HA was found to be 0.013 M. Determine the value of Ka for the weak acid. 1 2 NEXT Based on the given values, fill in the ICE table to determine concentrations of all reactants and products. HA(aq) H:O(1) H:O“(aq) A-(aq) + + Initial (M) 0.015 Change (M) -X +x +x Equilibrium (M) 0.015 - x +x +x
0.030 moles of a weak acid, HA, was dissolved in 2.0 L of water to form a solution. At equilibrium, the concentration of HA was found to be 0.013 M. Determine the value of Ka for the weak acid. 1 2 NEXT Based on the given values, fill in the ICE table to determine concentrations of all reactants and products. HA(aq) H:O(1) H:O“(aq) A-(aq) + + Initial (M) 0.015 Change (M) -X +x +x Equilibrium (M) 0.015 - x +x +x
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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This is the answer I got for this problem, but It's saying I have the change in concentration wrong. I don't understand what the help prompt is saying here.
I also have Ka = 2x/(0.015-x) = 0.0003
Is that correct? It didn't fit on the snip I took, sorry.
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