5.

5. HOBr(aq) ↔ H+(aq) + OBr–(aq) Ka = 2.3x10–9

Hypobromous acid, HOBr, is a weak acid that dissociates in water, as represented by the equation above.

(a) Calculate the value of [H+] in an HOBr solution that has a pH of 4.95.

(b) Write the equilibrium constant expression for the ionization of HOBr in water, then calculate the concentration of HOBr(aq) in an HOBr solution that has [H+] equal to 1.8x10–5 M.

(c) Would the Ka for HOI be greater than, less than or equal to that of HOBr? Explain your reasoning.

Expert Solution

queen_honey_blossom answered 3 years ago

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