Question

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.

Answer 1

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?

Answer 2

constant
sorry :(

Answer 3

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^+).

Answer 4

I got the pH how do i turn that into (H+)

Answer 5

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 10^x button and up will pop 5.01187 x 10^-5. Of course you would round that to 5.01 x 10^-5 M