Question
In: Chemistry
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.
Solutions
queen_honey_blossom answered 1 year agoRelated Solutions
STARS ARR_AUG 5 160 5 173 5 174 5 225 5 195 5 136 5 114...
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