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3. Using Equation 18-5 and the pKIn value you found, determine the ratio [In-]/[HIn] at the pH in the middle of the range you studied (i.e., pH = 6.0 for bromocresol purple, pH = 7.0 for bromothymol blue, and pH = 7.7 for phenol red). Show how you accomplished this. Draw a conclusion about the relative concentrations of In- and HIn at pH values very close to the pKIn of the indicator.

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maranathaoye

Answer:

Explanation:

[In-]/ [HIn] = [KIn]/ [H3O+]

Use the  values of  PkIN = 7.1,

KIN=8.14E-8, and

[H3O+] = 10E-7

To plug them in as follows:

(8.14E-8)/ (10E-7) = 0814, thus shows a  very close  ratio 1 to 1

The relative concentrations of In- and HIn at pH values are very close to the pKIn of the indicator to a close-ratio of 1 to 1

What is Bromocresol purple?

When A brominated acid dye of the sulfonephthalein series derived from ortho-cresol is acquired as a pinkish crystalline powder and also it is used as an acid-base indicator.

[In-]/ [HIn] is = [KIn]/ [H3O+]

Then we Use the values of PkIN is = 7.1,

Now, KIN is = 8.14E-8, and

[H3O+] is = 10E-7

Now To plug them in as follows:

Therefore, (8.14E-8)/ (10E-7) = 0814, thus shows a very close ratio 1 to 1

Find more information about Bromocresol purple here:

https://brainly.com/question/26387505

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