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HOBr(aq) ----> H+(aq) + OBr-(aq) Ka = 2.3 x 10-9

1. Hydrobromous acid, HOBr, is a weak acid that disassociates in water, as represented by the equation above.

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

(b) Write the equilibrium constant expression for the ionization of the HOBr in water, then calculate the concentration of the HOBR(aq) in an HOBr solution that has [H+] equal to 1.8 x 10-5 M.

(c) A solution of Ba(OH)2 is titrated into a solution of HOBr.
(i) Calculate the volume of 0.115 M Ba(OH)2 (aq) needed to reach the equivalence point when titrated into a 65.0 mL sample of 0.146 M HOBr(aq)
(ii) Indicate whether the pH at the equivalence point is less than 7, equal to 7 or greater than 7. Explain.

(d) Calculate the number of moles of NaOBr(s) that would have to be added to 125 mL of 0.160 M HOBr to produce a buffer solution with [H+] = 5.00 x 10-9 M. Assume that volume change is negligible.

(e) HOBr is a weaker acid than HBrO3. Account for this fact in terms of molecular structure.

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HOBr(aq) ↔ H+(aq) + OBr-(aq) Ka = 2.3 x 10-9

1. Hydrobromous acid, HOBr, is a weak acid that disassociates in water, as represented by the equation above.

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

(b) Write the equilibrium constant expression for the ionization of the HOBr in water, then calculate the concentration of the HOBR(aq) in an HOBr solution that has [H+] equal to 1.8 x 10-5 M.
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