The answer for this issue is:
The chemical equation is: HBz + H2O <- - > H3O+ + Bz-
Ka = 6.4X10^-5 = [H3O+][Bz-]/[HBz]
Let x = [H3O+] = [Bz-], and [HBz] = 0.5 - x.
Accept that x is little contrasted with 0.5 M. At that point,
Ka = 6.4X10^-5 = x^2/0.5
x = [H3O+] = 5.6X10^-3 M
pH = 2.25
(x is without a doubt little contrasted with 0.5, so the presumption above was OK to make)
Answer:
[IBr] = 0.049 M.
Explanation:
Hello there!
In this case, according to the balanced chemical reaction:
It is possible to set up the following equilibrium expression:
![K=\frac{[IBr]^2}{[I_2][Br_2]} =0.0110](https://tex.z-dn.net/?f=K%3D%5Cfrac%7B%5BIBr%5D%5E2%7D%7B%5BI_2%5D%5BBr_2%5D%7D%20%3D0.0110)
Whereas the the initial concentrations of both iodine and bromine are 0.50 M; and in terms of 
(reaction extent) would be:
Which can be solved for to obtain two possible results:
Whereas the correct result is 0.0245 M since negative results does not make any sense. Thus, the concentration of the product turns out:
![[IBr]=2x=2*0.0249M=0.049M](https://tex.z-dn.net/?f=%5BIBr%5D%3D2x%3D2%2A0.0249M%3D0.049M)
Regards!
The strength of an Arrhenius base determines percentage of ionization of base and the number of OH⁻ ions formed.
Strong base completely ionize in water and gives a lot of hydroxide ions (OH⁻), for example sodium
hydroxide: NaOH(aq) → Na⁺(aq)
+ OH⁻(aq).
Weak base partially ionize in water and gives a few hydroxide ions (OH⁻), for example ammonia: NH₃ + H₂O(l) ⇄ NH₄⁺(aq) + OH⁻(aq).
