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.

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Expert Solution

queen_honey_blossom answered 2 years ago
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