0.10- Formic acid, HFor, has a Kvalue of 1.8 x 10-4. You need to prepare 150 mL of a buffer hav- ing a pH of 3.25 from 0.10 M HFor solution and a 0.10 M NaFor solution. How many mL of NaFor and HFor should be mixed to make the desired buffer? -4 -
0.10- Formic acid, HFor, has a Kvalue of 1.8 x 10-4. You need to prepare 150 mL of a buffer hav- ing a pH of 3.25 from 0.10 M HFor solution and a 0.10 M NaFor solution. How many mL of NaFor and HFor should be mixed to make the desired buffer? -4 -
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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Question 4 and 5
![201×の-91X カbム
CN-] xHF]
[NOH
66.304 x10-6
%3D
メx7
%3D
メ-01
4. Formic acid, HFor, has a K value of 1.8 × 10-4. You need to prepare 150 mL of a buffer hav-
ing a pH of 3.25 from 0.10 M HFor solution and a 0.10 M NaFor solution. How many mL
of NaFor and HFor should be mixed to make the desired buffer?
mL HFO.
mL NaFor
LsatE]
(NaFor]
CHFOR]
5. When five drops of 0.10 M NaOH were added to 20 mL of the buffer in question 4, the pH
went from 3.25 to 3.31. Write a net ionic equation to explain why the pH did not go up more
than this.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fddc1ec45-98e4-4675-b97b-277ce49cbbd0%2F97b3b5bd-6d6c-48a0-9441-a288d9f80f0e%2Fx7am3v.jpeg&w=3840&q=75)
Transcribed Image Text:201×の-91X カbム
CN-] xHF]
[NOH
66.304 x10-6
%3D
メx7
%3D
メ-01
4. Formic acid, HFor, has a K value of 1.8 × 10-4. You need to prepare 150 mL of a buffer hav-
ing a pH of 3.25 from 0.10 M HFor solution and a 0.10 M NaFor solution. How many mL
of NaFor and HFor should be mixed to make the desired buffer?
mL HFO.
mL NaFor
LsatE]
(NaFor]
CHFOR]
5. When five drops of 0.10 M NaOH were added to 20 mL of the buffer in question 4, the pH
went from 3.25 to 3.31. Write a net ionic equation to explain why the pH did not go up more
than this.
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