HBrO ==> H^+ + BrO^-
Ka = 2.0 x 10^-9 = (H^+)BrO^-)/(HBrO)
Set up ICE chart, substitute and solve.
Equilibrium:
H^+ = X
BrO = X
HBrO = 0.15 -X
(The X is negligible compared to 0.15)
So we have X^2/0.15 = 2.0 x 10^-9
Solve for X and I get 1.73 x 10^-5
pH = -log(H^+) = -log(1.73 x 10^-5) = -(-4.76) = +4.76 :-).
a titration is done with with 25.00 mL of a .1500 M HBrO solution is titrated with .1025 M NaOH. Ka for HBrO= 2.0 x10^-9. The pH of HBrO before the titration begins is approx. closest to which of the following
a. 0.82
b. 4.76
c. 8.70
d 10.60
e. 13.01
the answer is b 4.76 but how?
1 answer