18. Calculate the pH of the solution formed when 12.0g of sodium fluoride (0.286mol) (NaF, FW- 41.99 is added to 500.ml of an 0.500M solution of hydrofluoric acid (HF). We start off with a weak acid in solution. When we add the salt of its conjugate base, we make a buffer solution, and the Na" is a spectator ion: The pka for HF is 6.4 x 10 (Acid]-0.500M nase -(Base] = Vaotn mol 0.286mol 12.0g x - 0.286mol 500.x 10-L - 0.572 M 41.99g K (64 x 10) Complete the problem by calculating the pH of the buffer solution.

18. Calculate the pH of the solution formed when 12.0g of sodium fluoride (0.286mol) (NaF, FW- 41.99 is added to 500.ml of an 0.500M solution of hydrofluoric acid (HF). We start off with a weak acid in solution. When we add the salt of its conjugate base, we make a buffer solution, and the Na" is a spectator ion: The pka for HF is 6.4 x 10 (Acid]-0.500M nase -(Base] = Vaotn mol 0.286mol 12.0g x - 0.286mol 500.x 10-L - 0.572 M 41.99g K (64 x 10) Complete the problem by calculating the pH of the buffer solution.

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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PH of buffer solution?

18. Calculate the pH of the solution formed when 12.0g of sodium fluoride (0.286mol) (NaF, FW-
41.99 is added to 500.mL of an 0.500M solution of hydrofluoric acid (HF).
We start off with a weak acid in solution. When we add the salt of its conjugate base, we make a
buffer solution, and the Na" is a spectator ion: The pKa for HF is 6.4 x 10
[Acid] 0.500M
nease
(Base]=
Vaotn
0.286mol
500.x 10L
12.0g x
41.99g
mol
- 0.286mol
- 0.572 M
K (6.4 x 10)
Complete the problem by calculating the pH of the buffer solution.

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Use the References to access important values if needed for this question. The pH of an aqueous solution of 0.530 M aniline (a weak base with the formula CHNH2) is Submit Answer 5 question attempts remaining
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