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

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answerhappygod
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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-]
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