0.00130 molk (CH2)NH*POH = - log (u.8x 10-5)- log (0.8x 10-s)4.50 + 20.0D = 214.50 x 10-3Llog0.00110 mols (CH)N24.50 x 10-3L

This is Henderson equation for buffer. Volume is cancel out as it is same

Answered: 0.00130 molk cry]g NH* = Hod - log…

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0.00130 molk cry]g NH* = Hod - log (u.8x 10-s) 4.50 + 20.0D = 24.50 x 1l0°3L • log 0. 00110 mols (CH)N 24.50 x 10- 3L

Introduction to General, Organic and Biochemistry

11th Edition

ISBN:9781285869759

Author:Frederick A. Bettelheim, William H. Brown, Mary K. Campbell, Shawn O. Farrell, Omar Torres

Publisher:Frederick A. Bettelheim, William H. Brown, Mary K. Campbell, Shawn O. Farrell, Omar Torres

Chapter6: Solutions And Colloids

Section: Chapter Questions

Problem 6.29P

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Acid Base Titration

An acid-base titration is the method of quantitative analysis for determining the concentration of an acid and base by exactly neutralizing it with a standard solution of base or acid having known concentration.

Question

0.00130 molk (CH2)NH*
POH = - log (u.8x 10-5)
- log (0.8x 10-s)
4.50 + 20.0D = 214.50 x 10-3L
log
0.00110 mols (CH)N
24.50 x 10-3L

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