In the titration of 25.00 mL of diprotic acid butyric acid (let's call it H2B) 0.100 M with KOH 0.100M the pH between th

Business, Finance, Economics, Accounting, Operations Management, Computer Science, Electrical Engineering, Mechanical Engineering, Civil Engineering, Chemical Engineering, Algebra, Precalculus, Statistics and Probabilty, Advanced Math, Physics, Chemistry, Biology, Nursing, Psychology, Certifications, Tests, Prep, and more.
Post Reply
answerhappygod
Site Admin
Posts: 899603
Joined: Mon Aug 02, 2021 8:13 am

In the titration of 25.00 mL of diprotic acid butyric acid (let's call it H2B) 0.100 M with KOH 0.100M the pH between th

Post by answerhappygod »

In the titration of 25.00 mL of diprotic acid butyric acid
(let's call it H2B) 0.100 M with KOH 0.100M the pH between the
first equivalence point and the second equivalence point would
be
a. pH=pK a1 + log [HB-]/[H2B]
b. pH = ½(pKa1+pKa2)
c. pH=pK a2 + log [H2B]/[ B-2]
d. pH=pK a2 + log [B-2]/[ HB-]
Join a community of subject matter experts. Register for FREE to view solutions, replies, and use search function. Request answer by replying!
Top
Post Reply

Return to “Archived Questions & Answers”