What is the definition of kb and the ph .10Tedu.COQuestion 12 of 18Determine the pH at the equivalence point in the titration of 50.0 mL of 0.300 MCH;COOH with 0.300 M NaOH. The value of Ka for CH3COOH is 1.8 × 10-5.1234NEXTUse the table below to determine the moles of reactant and product after the reaction of the acidand base. You can ignore the amount of liquid water in the reaction.CH;COOH(aq) +ОН (ад)H2O(1)+ CH3COO (aq)Before (mol)Change (mol)After (mol)5 RESET0.300-0.300+x+0.300 + x0.300 - x15.0-15.00.0300-0.03000.150-0.1500.0150-0.015050.0-50.0 Question 12 of 18Determine the pH at the equivalence point in the titration of 50.0 mL of 0.300 MCH3COOH with 0.300 M NaOH. The value of Ka for CH3COOH is 1.8 x 10-5.34NEXTPREVBased on the result of the acid-base reaction, set up the ICE table in order to determine theunknown.CH;COO"(aq) +H;O(1)ОН (ад)+ CH;COOH(aq)Initial (M)Change (M)Equilibrium (M)5 RESET0.3000.03000.1500.01500.0200+x11110.300 + x0.300 - x0.0300 + x0.0300 - x0.150 + x0.150 - x0.0150 + x-X0.0150 - x0.0200 + x0.0200 - x

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Determine the pH at the equivalence point in the titration of 50.0 mL of 0.300 M CH;COOH with 0.300 M NaOH. The value of Ka for CH3COOH is 1.8 x 10-5.

Chemistry: Principles and Practice

3rd Edition

ISBN:9780534420123

Author:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward Mercer

Publisher:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward Mercer

Chapter16: Reactions Between Acids And Bases

Section: Chapter Questions

Problem 16.51QE

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Question

What is the definition of kb and the ph

Question 12 of 18
Determine the pH at the equivalence point in the titration of 50.0 mL of 0.300 M
CH3COOH with 0.300 M NaOH. The value of Ka for CH3COOH is 1.8 x 10-5.
3
4
NEXT
PREV
Based on the result of the acid-base reaction, set up the ICE table in order to determine the
unknown.
CH;COO"(aq) +
H;O(1)
ОН (ад)
+ CH;COOH(aq)
Initial (M)
Change (M)
Equilibrium (M)
5 RESET
0.300
0.0300
0.150
0.0150
0.0200
+x
1111
0.300 + x
0.300 - x
0.0300 + x
0.0300 - x
0.150 + x
0.150 - x
0.0150 + x
-X
0.0150 - x
0.0200 + x
0.0200 - x

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We have :

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Number of Moles :

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Base Dissociation constant, Kb

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Hydroxide Ion Concentration :

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Calculation for pOH

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