Please don't provide handwritten solution ..... 4For H3PO4. Ka1 = 7.3 × 10-3, K₂2= 6.2 x 10-8, and K₂3=4.8 x 10-13. What is theconcentration of the HPO42- in a phosphoric acid solution whose initial acidconcentration was 0.12 M?4.8 x 10-13 M8.6 x 10-5 M6.2 x 10-8 MO 7.3 × 10 ³ MNone of these choices are correct.

Answered: For H3PO4, Ka1 = 7.3 x 10³. Ka2 = 6.2 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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Please don't provide handwritten solution .....

4
For H3PO4. Ka1 = 7.3 × 10-3, K₂2= 6.2 x 10-8, and K₂3=4.8 x 10-13. What is the
concentration of the HPO42- in a phosphoric acid solution whose initial acid
concentration was 0.12 M?
4.8 x 10-13 M
8.6 x 10-5 M
6.2 x 10-8 M
O 7.3 × 10 ³ M
None of these choices are correct.

Expert Solution

Step 1: Determine the concentration of HPO4 2- from the given data:

Given,

Initial, [ H₃ ] = 0.12 M

$H subscript 3 P O subscript 4 space left parenthesis a q right parenthesis end subscript space plus space H subscript 2 O subscript left parenthesis l right parenthesis end subscript space left right double arrow space H subscript 2 P O subscript 4 to the power of minus subscript left parenthesis a q right parenthesis end subscript space plus thin space H subscript 3 O to the power of plus subscript left parenthesis a q right parenthesis end subscript space space space space space space K subscript a 1 end subscript space equals space 7.3 space x space 10 to the power of negative 3 end exponent H subscript 2 P O subscript 4 to the power of minus subscript left parenthesis a q right parenthesis end subscript space plus space H subscript 2 O subscript left parenthesis l right parenthesis end subscript space left right double arrow space H P O subscript 4 to the power of 2 minus end exponent subscript left parenthesis a q right parenthesis end subscript space plus thin space H subscript 3 O to the power of plus subscript left parenthesis a q right parenthesis end subscript space space space space space space K subscript a 2 end subscript space equals space 6.2 space x space 10 to the power of negative 8 end exponent H P O subscript 4 to the power of 2 minus end exponent subscript left parenthesis a q right parenthesis end subscript space plus space H subscript 2 O subscript left parenthesis l right parenthesis end subscript space left right double arrow space P O subscript 4 to the power of 3 minus end exponent subscript left parenthesis a q right parenthesis end subscript space plus thin space H subscript 3 O to the power of plus subscript left parenthesis a q right parenthesis end subscript space space space space space space space space K subscript a 3 end subscript space equals space 4.8 space x space 10 to the power of negative 13 end exponent$

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