Calculate the value of [H3O+] in a 0.01 M HOBr solution. Ka = 2.5E-9

Question

I'm having a problem with just writing the equilibrium expression.
do you add H2O to the HOBr?

HOBr + H2O <---> H3O+ + OBr-

but then you could say that because HOBr = .01 M, then 0Br- = .01 M, and H3O+ would equal the Ka. but that is not the answer...

Answer 1

Yes, technically, one adds water but many people omit the water.
HOBr + H2O ==> H3O^+ + OBr^- OR
HOBr ==> H^+ + OBr^-

Ka = 2.5E-9 = (H3O^+)(OBr^-)/(HOBr)

What you are doing wrong is you are assuming the ionization is 100% and it isn't (otherwise the Ka would be infinity). A solution 0.01 M in HCl will be 0.01M in H^+ and 0.01M in Cl^-; but that is because HCl is 100% ionized. HOBr, having a Ka of 2.5E-9 isn't anywhere close to 100% and you cannot do it this way. You set up an ICE chart, substitute into Ka expression, and solve for the unknown.
initial:
HOBr = 0.01
H^+ = 0
OBr^- = 0

change:
H^+ = +x
OBr^- = +x
HOBr = -x

equilibrium:
H^+ = +x
OBr^- = +x
HOBr = 0.01-x
Substitute into Ka expression above and solve for x.

Answer 2

thanks!!!