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Calculate the value of the acid-dissociation constant. Ka=?
The pH of 0.039M hypobromous acid (HOBr) is 5.06.
equilibrium equation for the dissociation of HOBr.
HOBr(aq)+H2O(l)⇌H3O+(aq)+OBr−(aq)
Not sure how to calc the value of Ka
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- steve_geo1Lv 76 years ago
HOBr ===> H+ + OBr-
Ka = [H+][OBr-]/[HOBr]
pH = -Log[H+], so +Log[H+] = -5.06
Antilog(-5.06) = 8.71 x 10^-6 = [H+] (Calculator)
Then [OBr-] = 8.71 x 10^-6 also.
Ka = (8.71x10^-6)(8.71x10^-6)/(0.039) = 1.94 x 10^-9
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