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.
..........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.
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?
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.
Thanks so much! I get it now!
To calculate the pH at the equivalence point in part c(ii), we need to consider the reaction between the strong base, Ba(OH)2, and the weak acid, HOBr.
The balanced equation for the reaction is:
HOBr (aq) + OH- (aq) → OBr- (aq) + H2O (l)
Since this is an acid-base reaction, we can use the equilibrium expression for the reaction:
Kb = ([OBr-][H2O])/[HOBr]
Given that Kb = 4.3 x 10^-6, we can substitute the concentrations of the species at the equivalence point into the equation. At the equivalence point, all of the HOBr will react, so the concentration of HOBr is 0 M. The concentration of OBr- can be determined from the balanced equation, which indicates a 1:1 mole ratio of HOBr to OBr-.
So, at the equivalence point, the concentration of OBr- is equal to the concentration of Ba(OH)2 that was added. In this case, it is 0.115 M.
Plugging these values into the Kb expression:
4.3 x 10^-6 = ([0.115][H2O])/0
Simplifying the equation, we find:
[H2O] = 0
At the equivalence point, the concentration of water is equal to the concentration of HOBr, as all of the HOBr has reacted. Therefore, at the equivalence point, [H2O] = [HOBr] = 0.146 M.
Since HOBr is a weak acid, we can use the equation pH = -log[H+]. Therefore, to find the pH at the equivalence point, we need to calculate the concentration of H+.
From the equation for the dissociation of HOBr, we know that, at equilibrium, [H+] = [OBr-].
Therefore, at the equivalence point, [H+] = [OBr-] = 0.146 M.
Taking the negative logarithm of [H+], we find:
pH = -log(0.146) = 0.834
Therefore, the pH at the equivalence point is approximately 0.834, or 10.79 (rounded to two decimal places).
Remember to check if the value of the pH is greater than 7 because we are titrating a weak acid with a strong base, which will likely result in a basic solution. In this case, the pH is greater than 7, confirming that the solution is basic at the equivalence point.