Substitute in the Henderson equation and solve for x : [CH₂COO-] [CH₂COOH] pH = pKa + log 4.93=4.76 + log (40+2x) (60-2x) (40+2x) (60-2x) 0.17 = log (40+2x) €1.479 (60-2x) On solving, x = 9.83 So, volume of NaOH required = 9.83 mL

Algebra and Trigonometry (6th Edition)
6th Edition
ISBN:9780134463216
Author:Robert F. Blitzer
Publisher:Robert F. Blitzer
ChapterP: Prerequisites: Fundamental Concepts Of Algebra
Section: Chapter Questions
Problem 1MCCP: In Exercises 1-25, simplify the given expression or perform the indicated operation (and simplify,...
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can someone explain how the circled value was calculated. Im not sure where the number came from? I see the log dissapeard but Im not sure how that happened. thx for explaining

Substitute in the Henderson equation and solve for x :
pH = pKa + log
[CH₂COO-]
[CH₂COOH]
(40+2x)
(60-2x)
4.93=4.76 + log
(40+2x)
(60-2x)
0.17 = log
(40+2x) 1.479
(60-2x)
On solving, x = 9.83
So, volume of NaOH required = 9.83 mL
Transcribed Image Text:Substitute in the Henderson equation and solve for x : pH = pKa + log [CH₂COO-] [CH₂COOH] (40+2x) (60-2x) 4.93=4.76 + log (40+2x) (60-2x) 0.17 = log (40+2x) 1.479 (60-2x) On solving, x = 9.83 So, volume of NaOH required = 9.83 mL
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