An indicator HIn has K_a = 1 *10^-⁸. At pH = 6.0, what is the ratio HIn/In^-?

Name: An indicator HIn has K a = 1 *10 - 8 . At pH = 6.0, what is the ratio HIn/In - ?
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Question

Quizplus · Accepted Answer

The pH is 6.0, and the pKa of the indicator is 8.0. Using the Henderson-Hasselbalch equation: pH = pKa + log([In-]/[HIn]), we find that log([In-]/[HIn]) = 6 - 8 = -2. Therefore, [In-]/[HIn] = 10^-2, which means [HIn]/[In-] = 100/1.

An Indicator HIn Has Ka = 1 *10-8

An Indicator HIn Has K_a = 1 *10^-⁸