What is the pOH of 2.26 M Ba(OH)2?
Ba(OH)2 ==> Ba^+2 + 2OH^-
(OH)^- = 2 x 2.26 M.
pOH = - log(OH^-) = ??
yes
To find the pOH of 2.26 M Ba(OH)2, we first need to determine the concentration of hydroxide ions (OH^-) in the solution.
Since each Ba(OH)2 molecule dissociates into one Ba^+2 ion and two OH^- ions, we can conclude that the concentration of OH^- ions is twice the concentration of Ba(OH)2. Therefore:
(OH)^- = 2 x 2.26 M = 4.52 M
Now, we can calculate the pOH using the formula:
pOH = -log[OH^-]
Plugging in the value, we have:
pOH = -log[4.52] ≈ 0.343
So, the pOH of a 2.26 M Ba(OH)2 solution is approximately 0.343.