The dissociation for a weak base BOH in water found to be 1.25 x 10^-6. What is the concentration of H+ in a 3.2M soluton of BOH.
BOH <==> B^+ + OH^-
Kb = (B^+)(OH^-)/(BOH) = 1.25 x 10^-6
Have you been introduced to the ICE chart?
Initial = I
change = C
equilibrium = E
Initial concns (before any dissociation):
(BOH) = 3.2 M
(B^+) = 0
(OH^-) = 0
change in concns (after dissociation):
(B^+) = y
(OH^-) = y
(BOH) = -y
equilibrium concns:
(B^+) = 0 + y = y
(OH^-) = 0 + y = y
(BOH) = 3.2 - y = 3.2-y
Plug the equilibrium values into the Kb expression and solve for y. You will get a quadratic which you can solve with the quadratic formula OR you can make the simplifying assumption and see if that will work. At any rate, y = (OH^-) so you take the - log to get pOH, then subtract from 14 to get pH, then convert to (H^+).
Do you mean the dissociation for a weak base is 1.25 x 10^-5 or the dissociation constant is 1.25 x 10^-5?
constant
sorry :(
I got the pH how do i turn that into (H+)
pH = -log(H^+)
Say pH = 4.3.
Then 4.3 = -log(H^+)
-4.3 = log(H^+)
Now enter -4.3 into your calculator, then hit the 10x button and up will pop 5.01187 x 10^-5. Of course you would round that to 5.01 x 10^-5 M
To find the concentration of H+ in a solution of BOH, we need to consider the dissociation of the weak base.
The dissociation of BOH can be represented by the following equation:
BOH ⇌ B+ + OH-
From the dissociation constant expression for a weak base:
Kw = [B+][OH-] / [BOH]
where Kw is the ionization constant for water, which is 1.0 x 10^-14 at 25°C.
Since the concentration of H+ and OH- in pure water at 25°C is equal (10^-7 M), we can approximate the concentration of OH- in the solution of BOH to be 1.25 x 10^-6.
Hence, we can rewrite the equation as:
1.0 x 10^-14 = [B+][1.25 x 10^-6] / [BOH]
Rearranging the equation to solve for the concentration of H+:
[H+] = (1.0 x 10^-14) / (1.25 x 10^-6) x [BOH]
Given that the concentration of BOH is 3.2 M, we can substitute the values:
[H+] = (1.0 x 10^-14) / (1.25 x 10^-6) x 3.2
Calculating this expression will give us the concentration of H+ in the solution of BOH.