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

Question

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.

Answer 1

..........OBr^- + HOH-->HOBr + OH^-
initial..0.0893..........0......0
change....-x.............x.......x
equil...0.903-x..........x.......x

Kb for OBr^- = Kw/Ka for HOBr = (HOBr)(OH^-)/(OBr^-)
Substitute and solve for x = (OH^-) and convert to pH.

Answer 2

Hi Dr. Bob,

When I do that, I get 11.30 as the pH, although I don't think that's it. Did I miss something?

Answer 3

Maybe and maybe not. That .903 is not right. I remember typing 0.0893
Using 0.0893 I get 10.79. Your Kb is ok. That 0.0893 comes from [(65mL*0.146M/(65mL+41.3mL)]. Sorry I usually proof my responses to catch those typos; I didn't proof this one.
I calculated it using 0.903 and obtained 11.3 also.

Answer 4

Thanks so much! I get it now!