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Hypobromous acid (HOBr) is produced in the blood and has properties that combat pathogens. It is also used in bleaches and other cleaning products. The following is a reversible reaction:

HOBr → H+ + OBr–

a. Write the equation you would use to calculate the Ka of this reaction.

b. If [H+] at equilibrium is 9. 1 × 10–6, what is the Ka of the reaction, given a final HOBr concentration

of 0. 33 M?

Respuesta :

pstnonsonjoku

From the calculations in the question, we have that the dissociation constant is 2.5 * 10^-10.

What is Ka?

The term Ka has to do with the acid dissociation constant and it is obtained from; Ka = [H+] [OBr–]/[HOBr]

Now, we are told that the hydrogen ion concentration at equilibrium is 9. 1 × 10–6 and the HOBr concentration is 0.33 M, it the follows that;

Ka = (9. 1 × 10–6)^2/ 0.33

Ka = 2.5 * 10^-10

Learn more about equilibrium: https://brainly.com/question/10038290

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