Calculate [H3O+] and [OH-] in 0.18M hydrobromic acid, HOBr(aq), having a K of 2.0x10^-9?
I got 9.0x10^-8
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.
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
#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.
To calculate the concentration of [H3O+] and [OH-] in hydrobromic acid (HOBr(aq)), we need to consider the dissociation of the acid. Hydrobromic acid (HOBr) is a strong acid, which means it completely dissociates in water.
The dissociation equation for hydrobromic acid is as follows:
HOBr(aq) ⇌ H+(aq) + OBr-(aq)
Given that the hydrobromic acid has a concentration of 0.18 M, we can assume that the concentration of H+ and OBr- are also 0.18 M.
Since H+ and OH- ions are present in water in equal amounts due to the auto-ionization of water, we can calculate [OH-] as follows:
[H3O+][OH-] = Kw
Where Kw is the ionization constant of water and has a value of 1.0 x 10^-14 at room temperature.
[H3O+][OH-] = 1.0 x 10^-14
Since [H3O+] and [OH-] are equal in a neutral solution, we can assign them both the variable x. This gives us the equation:
x * x = 1.0 x 10^-14
Solving this equation, we find that x = 1.0 x 10^-7. Therefore, [H3O+] and [OH-] both have a concentration of 1.0 x 10^-7 M in neutral water.
Now, in the case of hydrobromic acid, the concentration of [H3O+] is higher due to the additional H+ ions provided by the dissociation of the acid.
In this case, [H3O+] would be the initial concentration of hydrobromic acid (0.18 M) added to the concentration of H+ ions formed by dissociation. Since hydrobromic acid is a strong acid and completely dissociates, the concentration of H+ ions will be the same as the initial concentration of the acid.
Therefore, [H3O+] = 0.18 M.
Since [H3O+] and [OH-] are inversely related, we can use the equation [H3O+][OH-] = Kw to find [OH-]:
(0.18 M) * [OH-] = 1.0 x 10^-14
Solving for [OH-], we find that [OH-] = 5.6 x 10^-14.
Therefore, the concentration of [H3O+] in 0.18 M hydrobromic acid is 0.18 M, and the concentration of [OH-] is 5.6 x 10^-14 M.