Answer to Question #9666 in Biochemistry for Meri

Question

1. 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?

Answer 1

a. Ka=[H+][OBr -] / [HOBr]
b. HOBr → H+ + OBr–
in equiation
[H+]=[OBr-],so
Ka=[H+]^2 /[HOBr]=(9.1 × 10–6)^2 /0.33 =2,51*10^-10