Question: Two drops of indicator HIn (Ka = 1.0 109), where HIn is yellow and In- is blue, are placed in 100.0 mL of 0.10

Two drops of indicator HIn (Ka = 1.0 × 10–9), where HIn is yellow and In- is blue, are placed in 100.0 mL of 0.10 M HCl.

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HIn In H K a 10 a In a very acid solution the HIn form dominates so the solution will be yellow ... View full answer

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