Part A Calculate the pH of a buffer solution that is 0.265 molal in CH3COOH (K₂ = 1.75 x 10-5) and 0.385 molal in CH3COONa using the Davies equation to calculate + at 298.15 K. Express your answer using two decimal places. pH = , Templates Symbols undo redo reset keyboard shortcuts 'help' help! AS 1 r Submit Request Answer Part B What pH value would you have calculated if you had assumed that 7+ = 1? Express your answer using two decimal places. pH = help r Templates Symbols undo redo reset keyboard shortcuts help ASA 1 Submit Request Answer

Part A Calculate the pH of a buffer solution that is 0.265 molal in CH3COOH (K₂ = 1.75 x 10-5) and 0.385 molal in CH3COONa using the Davies equation to calculate + at 298.15 K. Express your answer using two decimal places. pH = , Templates Symbols undo redo reset keyboard shortcuts 'help' help! AS 1 r Submit Request Answer Part B What pH value would you have calculated if you had assumed that 7+ = 1? Express your answer using two decimal places. pH = help r Templates Symbols undo redo reset keyboard shortcuts help ASA 1 Submit Request Answer

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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I have seen other "experts" answer this question but i'm not understanding the process. Please write out the steps and why. And please explain why we use 0.100-x and x.
+ Base/Acid Ratios in Buffers Just as pH is the negative logarithm of [H3O+]. PK₂ is the negative logarithm of Ka. PK₂ -log Ka The Henderson-Hasselbalch equation is used buffer solutions: calculate the pH of base DAGSK pH =pK₂ +log acid] Notice that the pH of a buffer has a value close to the PK₂ of the acid. differing only by the logarithm of the concentration ratio [base]/[acid]. The Henderson-Hasselbalch equation in terms of pOH and PKb is similar. acid [base pOH = pkt + log- Part A Acetic acid has a K₁ of 1.8 x 105. Three acetic acid/acetate buffer solutions, A, B, and C, were made using varying concentrations: A. [acetic acid] ten times greater than [acetate]. B. [acetate] ten times greater than [acetic acid], and c. [acetate] [acetic acid]. Match each buffer to the expected pH Drag the appropriate items to their respective bins. ▸ View Available Hint(s) [acetic acid] ten times greater than [acetate] pH = 3.74 [acetate] ten times greater than [acetic acid] pH = 4.74 Reset Help…
A student needs to prepare a buffer made from CH3COOH and CH3COOK with pH 5.074. If K₂ for CH3COOH is [CH, COOH] [CH₂COOT] is required? C 1.80 x 10-5, what ratio of [CH3COOH] [CH, COO Submit Answer $ 000 R F V = % 5 [Review Topics] [References] Use the References to access important values if needed for this question. Retry Entire Group 9 more group attempts remaining T G Cengage Learning | Cengage Technical Support FS ^ 6 B MacBook Air ---F6 Y H & 7 N ◄◄ F7 U J * 00 8 ►ll M - ( 9 K F9 O Save and Exit dele
7. Refer to the following table: Table 2-2. Change in pH of 0.2 M Acetate Buffer after addition of Acid or Base [Base]:[Acid] 0.005:0.195 Addition pH HC1 0.095 mol 3.1 HC1 0.075 0.025:0.175 3.9 HС1 0.05 НС1 0.025 0.05:0.15 4.2 0.075:0.125 4.5 0.1:0.1 0.125:0.075 4.7 NaOH 0.025 mol 4.9 NaOH 0.05 mol 0.15:0.05 5.2 NaOH 0.75 mol 0.175:0.025 5.5 NaOH 0.095 mol 0.195:0.005 6.3 a. Calculate the pH of a 0.2 M acetate buffer (a solution containing 0.1 mol L-' acetic acid and 0.1 mol L-1 sodium acetate), given that the pKa of acetic acid is 4.7. b. What would be the pH value after adding 0.05 mol of NaOH to 1 L of 0.2 mol L1 acetate buffer? c. Compare the pH value in (b) with that obtained after adding 0.05 mol NAOH to 1 L of water (a simple solution of 0.05 mol L-' NaOH). d. How much acid or base can be added to 1 L of the 0.2 M acetate buffer so that the pH does not change appreciably (i.e., what is its buffer range)?
Calculate the pH of 25 mL of 0.175 M CH3COOH with 30 mL of water added after the addition of 43.75 mL of 0.1 M NaOH (equivalence). Show work, please. This pertains to the weak acid strong base exercise.
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Calculate the volumes needed to create a 50 mL buffer solution with a target pH of 4.20. Use 0.10 M solutions from List E and List F. (Use HCOOH and HCOOK, and check if the work provided below is correct) HCOOH, HCOOK A CB он = рка+ List E 0.10 M HCOOH K = 1.8 x 10-4 0.10 M CH₂COOH K₁ = 1.8 x 10-5 0.10 M NH OH K=1.8 x 10-5 0.10 M CHẠNH, K = 4.4 x 10-4 ↓ List F 0.10 M CH,NH,CI 0.10 M NHẠCI 0.10 M CH,COONa 0.10 M HCOOK 0.10 M HCOONa 0.10 M CH,COOK 4.20 3.74 + 2.88 = CHCOOK] Снсоон] 50mL, .050L X= vol HCOOK 050-XE VỚI HCCH [B] [A] 2.88= 2.56 [HOOOK] log pнссон] 2x (0.10M-X)/.05L (0-10M (050-x)/05L 0.1-2x 2x 0.1-2x 2.88 0.1-2x) = 2x x = 0.0371 0.050-x = 0.0129 371mL HCOOK 12.9m2 HcoH
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Hello, I don't know where I'm going wrong. I'm using the Henderson-Hasselbalch equation. I've attempted three time with no success. Could you break down this problem so I can spot where I am going wrong? Thank you. Screenshot of problem is attached.
Considering the K, values shown below, estimate the pH at which the the titration described in the procedure will reach the first equivalence point. Give your answer to two decimal places. Ka1 = 6.9 x 10-3 %3D K32 = 6.2 x 10-8 %3D Ka3 = 4.8 x 10-13 Hint: Have you looked at the discussion slides? Your Answer: Answer
help with part b please