In the titration of a 25.0-mL sample of 0.145 M HCOOH with 0.122 M NaOH, where Ka, HCOOH = 1.8x10-4 %3D O the volume of NaOH added at one-half the equivalence point is about 27.4 mL O the volume of NaOH added about 7.45 mL one-half the equivalence point is O the volume of NaOH added at one-half the equivalence point is about 12.6 mL O the volume of NaOH added at one-half the equivalence point is about 14.9 mL
In the titration of a 25.0-mL sample of 0.145 M HCOOH with 0.122 M NaOH, where Ka, HCOOH = 1.8x10-4 %3D O the volume of NaOH added at one-half the equivalence point is about 27.4 mL O the volume of NaOH added about 7.45 mL one-half the equivalence point is O the volume of NaOH added at one-half the equivalence point is about 12.6 mL O the volume of NaOH added at one-half the equivalence point is about 14.9 mL
Chemistry
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ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
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Chapter1: Chemical Foundations
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![In the titration of a 25.0-mL sample of 0.145 M HCOOH with 0.122 M
NaOH, where K, HCOOH = 1.8x10-4
%3D
the volume of NAOH added at one-half the equivalence point is
about 27.4 mL
O the volume of NaOH added at one-half the equivalence point is
about 7.45 mL
O the volume of NaOH added at one-half the equivalence point is
about 12.6 mL
O the volume of NaOH added at one-half the equivalence point is
about 14.9 mL](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F0adf00ea-41d4-473a-97c2-219afdd856b3%2Fe9ef7f3c-54c6-4158-ba2f-547672604d5d%2Fd3hitjo_processed.jpeg&w=3840&q=75)
Transcribed Image Text:In the titration of a 25.0-mL sample of 0.145 M HCOOH with 0.122 M
NaOH, where K, HCOOH = 1.8x10-4
%3D
the volume of NAOH added at one-half the equivalence point is
about 27.4 mL
O the volume of NaOH added at one-half the equivalence point is
about 7.45 mL
O the volume of NaOH added at one-half the equivalence point is
about 12.6 mL
O the volume of NaOH added at one-half the equivalence point is
about 14.9 mL
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