I don't think your one (1) answer is right for either H^+ or OH^-
Post your work and I'll look for the error.
Calculate [H3O+] and [OH-] in 0.18M hydrobromic acid, HOBr(aq), having a K of 2.0x10^-9?
I got 9.0x10^-8
3 answers
i got
K= (HOBr2)(OH)/(HOBr)
2.0x10^-9 = (x)(x)/0.18M
x= 11x10^-6
(OH-) =11x10^-6
(H3O)= 1.0x10^-14/11x10^-6 =9.0x10^-8
K= (HOBr2)(OH)/(HOBr)
2.0x10^-9 = (x)(x)/0.18M
x= 11x10^-6
(OH-) =11x10^-6
(H3O)= 1.0x10^-14/11x10^-6 =9.0x10^-8
#1. Hydobromic acid is HBr. That is a strong acid. HOBr, I assume, is correct since that is a weak acid but it is called hypobromous acid.
HOBr + H2O ==> H3O^+ + OBr^-
Ka = 2.0E-9 = (H3O^+)(OBr^-)/(HOBr)
(H3O^+) = x
(OBr^-) = x
(HOBr) = 0.18-x which can be called 0.18 without changing the result but it makes the math easier.
Solve for x = (H3O^+)
After you know (H3O^+), then
(H3O^+)(OH^-) = 1E-14 and solve for (OH^-).
I obtained approximately 2E-5 for H^+ which makes OH^- about 5E-10. You should recalculate more accurately than I did but my answers are close.
HOBr + H2O ==> H3O^+ + OBr^-
Ka = 2.0E-9 = (H3O^+)(OBr^-)/(HOBr)
(H3O^+) = x
(OBr^-) = x
(HOBr) = 0.18-x which can be called 0.18 without changing the result but it makes the math easier.
Solve for x = (H3O^+)
After you know (H3O^+), then
(H3O^+)(OH^-) = 1E-14 and solve for (OH^-).
I obtained approximately 2E-5 for H^+ which makes OH^- about 5E-10. You should recalculate more accurately than I did but my answers are close.