Question
Calculate the value of [H3O+] in a 0.01 M HOBr solution. Ka = 2.5E-9
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...
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...
Answers
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.
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.
thanks!!!
Related Questions
15ml of 0.325M NaOH is delivered into 35ml HOBr solution of unknown concentration.The pH of the fina...
HOBr (aq) <----> H+ (aq) + OBr- (aq), Ka = 2.3 x 10^-9
Hypobromous acid, HOBr, is a weak acid that...
Hypobromous acid, HOBr, is a weak acid that dissociates in water, as represented by the equation
b...
the equilibrium constant expression for the ionization of HOBr in water, then calculate
the conce...