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b.) Starting with a 0.55 M solution of H2A, you want to know the pH and the final concentrations of all ions present. You find the Ka values for the two dissociation steps: Ka1 = 1.5 x 10-4 Ka2 = 2.0 x 10-7 i.) The first step for this problem is to consider the first dissociation reaction on its own. Below, solve for the concentrations of the ions involved at the end of the first dissociation and the percent ionization. You should end up with 3 total concentrations.

b.) Starting with a 0.55 M solution of H2A, you want to know the pH and the final concentrations of all ions present. You find the Ka values for the two dissociation steps: Ka1 = 1.5 x 10-4 Ka2 = 2.0 x 10-7 i.) The first step for this problem is to consider the first dissociation reaction on its own. Below, solve for the concentrations of the ions involved at the end of the first dissociation and the percent ionization. You should end up with 3 total concentrations.

Chemistry
Chemistry
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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
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please refer to images and answer bi bii and biii

b.) Starting with a 0.55 M solution of H2A, you want to know the pH and the final concentrations of all ions present. You find the Ka values for the two dissociation steps: Ka1 = 1.5 x 10-4 Ka2 = 2.0 x 10-7 i.) The first step for this problem is to consider the first dissociation reaction on its own. Below, solve for the concentrations of the ions involved at the end of the first dissociation and the percent ionization. You should end up with 3 total concentrations.
Transcribed Image Text:b.) Starting with a 0.55 M solution of H2A, you want to know the pH and the final concentrations of all ions present. You find the Ka values for the two dissociation steps: Ka1 = 1.5 x 10-4 Ka2 = 2.0 x 10-7 i.) The first step for this problem is to consider the first dissociation reaction on its own. Below, solve for the concentrations of the ions involved at the end of the first dissociation and the percent ionization. You should end up with 3 total concentrations.
ii.) Now you've accounted for the ionization of the first proton. But what about the second step? You'll need to make another RICE chart to figure out what the final concentrations of all the ions will be. Think carefully about what your initial values should be. (Hint: instead of x, you might want to use y this time to make your work less confusing). Solve for all the final concentrations and the pH. You should end up with 4 total concentrations. iii.) If all went well, you should have found that y is approximately equal to Ka2. Why is this the case? (Look at your Ka2 expression and consider the size of y).
Transcribed Image Text:ii.) Now you've accounted for the ionization of the first proton. But what about the second step? You'll need to make another RICE chart to figure out what the final concentrations of all the ions will be. Think carefully about what your initial values should be. (Hint: instead of x, you might want to use y this time to make your work less confusing). Solve for all the final concentrations and the pH. You should end up with 4 total concentrations. iii.) If all went well, you should have found that y is approximately equal to Ka2. Why is this the case? (Look at your Ka2 expression and consider the size of y).
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