Problem: The indicator dinitrophenol is an acid with a Ka of 1.1 × 10 −4. In a 1.0 × 10 −4-M solution, it is colorless in acid and yellow in base. Calculate the pH range over which it goes from 10% ionized (colorless) to 90% ionized (yellow).

🤓 Based on our data, we think this question is relevant for Professor Atwood's class at UB.

FREE Expert Solution

(1) Setup the Ka expression

Ka = [C6H3NO22O-][H3O+]C6H3(NO2)2OH


(2) Plugging in the values of concentration to the Ka expression:

At 10% dissociation: 

1x1x10-4 (0.10) = 1.1x10-5

At 90% dissociation: 

1x1x10-4 (0.90) = 9x10-5

Calculating H3O+ for both dissociation:

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Problem Details

The indicator dinitrophenol is an acid with a Ka of 1.1 × 10 −4. In a 1.0 × 10 −4-M solution, it is colorless in acid and yellow in base. Calculate the pH range over which it goes from 10% ionized (colorless) to 90% ionized (yellow).

Frequently Asked Questions

What scientific concept do you need to know in order to solve this problem?

Our tutors have indicated that to solve this problem you will need to apply the Acid Base Indicators concept. If you need more Acid Base Indicators practice, you can also practice Acid Base Indicators practice problems.

What professor is this problem relevant for?

Based on our data, we think this problem is relevant for Professor Atwood's class at UB.

What textbook is this problem found in?

Our data indicates that this problem or a close variation was asked in Chemistry - OpenStax 2015th Edition. You can also practice Chemistry - OpenStax 2015th Edition practice problems.