If a solution of 0.5 M HOAc dissociates as follows: HOAc + H2O „® H3O+ + OAc-, what is the final [H3O+] in the solution? Ka for HOAc = 1.8„ª10-5. What is the pH of the above solution? What is the Kb for HOAc? not sure of the initial set up please help
...........HOAc + H2O ==> H3O^+ + OAc^-
I..........0.5M............0.......0
C..........-x..............x.......x
E.........0.5-x............x.......x
Substitute the E line into Ka expression and solve for x = (H3O^+) then convert to pH.
If you know Ka (and you do) then KaKb = Kw = 1E-14. You always know Kw; if you know either Ka or Kb the other can be calculated.
To find the final [H3O+] in the solution, we can use the given information of the dissociation reaction and the Ka value for HOAc.
First, let's look at the dissociation reaction:
HOAc + H2O ⇌ H3O+ + OAc-
From this reaction, we can see that 1 mole of HOAc produces 1 mole of H3O+. Therefore, the concentration of H3O+ is equal to the concentration of the dissociated acid.
In the given solution, the initial concentration of HOAc is 0.5 M. Since HOAc dissociates, the concentration of H3O+ will also be 0.5 M.
Therefore, the final [H3O+] in the solution is 0.5 M.
Moving on to finding the pH of the solution, pH is the logarithmic scale used to measure the concentration of H3O+ ions. It is calculated using the equation:
pH = -log[H3O+]
Given that the concentration of H3O+ is 0.5 M, we can calculate the pH as follows:
pH = -log(0.5) ≈ -(-0.301) ≈ 0.301
So, the pH of the solution is approximately 0.301.
Finally, to find the Kb for HOAc, we can use the equation:
Ka x Kb = Kw
where Kw is the ion product of water and is equal to 1.0 x 10^-14 at 25°C.
Given Ka for HOAc as 1.8 x 10^-5, we can rearrange the equation to solve for Kb:
Kb = Kw / Ka
Kb = (1.0 x 10^-14) / (1.8 x 10^-5) ≈ 5.56 x 10^-10
Therefore, the Kb for HOAc is approximately 5.56 x 10^-10.