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?
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