April 21, 2014

Homework Help: Chemistry

Posted by Confused on Tuesday, April 10, 2012 at 7:22pm.

HOBr (aq) <----> H+ (aq) + OBr- (aq), Ka = 2.3 x 10^-9
Hypobromous acid, HOBr, is a weak acid that dissociates in water, as represented by the equation.
(a) Calculate the value of [H+] in a solution of HOBr that has a pH of 4.95.
Ans: 1.1 * 10^-5 M

(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 1.8 x 10-5 M.
Ans: 0.14 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).
Ans: 41.3 mL

(ii) Calculate the pH at the equivalence point.
Ans: 10.79

I know how to do every part of the question EXCEPT part c(ii). I know that the pH is above 7, but how do I find 10.79? I found Kb = 4.3 * 10^-6, but I am unsure of what to do with that.

By the way, I'm sure about the answer to part i of c. And for reactions with weak acids being titrated by strong bases, the equiv. point can't be at pH 7; only strong acid and strong base titrations give that pH at the equiv. pt.

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