In the titration of a 25.0-mL sample of 0.115 M HNO, with 0.21 MNaOH, where K, HNO2 = 4.6x104%3DO The pH at the equivalence point is equal to 11.The pH at the equivalence point is < 5.O The pH at the equivalence point is > 7.The pH at the equivalence point is neutral.

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In the titration of a 25.0-mL sample of 0.115 M HNO, with 0.21 M NaOH, where K, HNO, = 4.6x104 %3D O The pH at the equivalence point is equal to 11. O The pH at the equivalence point is < 5. O The pH at the equivalence point is > 7. O The pH at the equivalence point is neutral.

In the titration of a 25.0-mL sample of 0.115 M HNO, with 0.21 M NaOH, where K, HNO, = 4.6x104 %3D O The pH at the equivalence point is equal to 11. O The pH at the equivalence point is < 5. O The pH at the equivalence point is > 7. O The pH at the equivalence point is neutral.

Chemistry: An Atoms First Approach

2nd Edition

ISBN:9781305079243

Author:Steven S. Zumdahl, Susan A. Zumdahl

Publisher:Steven S. Zumdahl, Susan A. Zumdahl

Chapter14: Acid- Base Equilibria

Section: Chapter Questions

Problem 110CP: A 0.400-M solution of ammonia was titrated with hydrochloric acid to the equivalence point, where...

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Concept explainers

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

In the titration of a 25.0-mL sample of 0.115 M HNO, with 0.21 M
NaOH, where K, HNO2 = 4.6x104
%3D
O The pH at the equivalence point is equal to 11.
The pH at the equivalence point is < 5.
O The pH at the equivalence point is > 7.
The pH at the equivalence point is neutral.

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