Calculate the pOH of a 0.410 M Ba(OH) solution.
8.62*10^-2
OH is 2x Ba(OH)2.
pOH = -log(OH^-)
Well, Ba(OH) is known as barium hydroxide. Personally, I've always found barium hydroxide quite basic. It’s the life of the party, always ready to neutralize those pesky acidic molecules.
Now, to calculate the pOH of this solution, we need to know the hydroxide concentration. The formula unit of barium hydroxide contains two hydroxide ions, so the concentration of hydroxide (OH-) would be twice the concentration of barium hydroxide (Ba(OH)2).
Given that the concentration of Ba(OH)2 is 0.410 M, we simply double that value to get the hydroxide concentration. So, the hydroxide concentration would be 0.820 M.
To calculate pOH, we use the formula pOH = -log[OH-]. Plugging in the hydroxide concentration, we get pOH = -log(0.820) ≈ 0.087.
So, the pOH of a 0.410 M Ba(OH) solution is approximately 0.087. It seems like this solution isn't too basic, but it still knows how to have a good time!
To calculate the pOH of a solution, we need to first determine the concentration of hydroxide ions (OH-) in the solution.
In the case of a Ba(OH) solution, each Ba(OH) molecule dissociates into one Ba 2+ ion and two OH- ions. Therefore, the concentration of OH- ions in the solution will be twice the concentration of Ba(OH).
Given that the concentration of Ba(OH) is 0.410 M, the concentration of OH- ions will be 2 * 0.410 M = 0.820 M.
Now, use the relationship between pOH and OH- concentration:
pOH = -log10 [OH-]
Substituting the given OH- concentration into the equation:
pOH = -log10 (0.820)
Calculating this value, the pOH of the solution is approximately 0.087.
pOH = -log([OH])
[OH]=concentration*# of hydroxide ions
pOH=-log(2*0.41)
pOH=8.62*10^-2