In this question, we need to calculate the pH of 0.00500 M of Ba(OH)2, and in order to solve this question, we will be using the pOH formula, which is:
pOH = -log [OH-]
The pOH is equal to minus log of the concentration of OH-, since this substance is a base, we have to find the value of OH-, let's add the values into the formula
But before that, we need to know what is the concentration of OH-, which we can see by the equation of dissociation of the base:
Ba(OH)2 -> Ba2+ + 2 OH-
In this formula we can see that we have twice the concentration of OH-, so, if we have a concentration of 0.00500 M for Ba2+, we will have an OH- concentration of 0.0100 M, and this is the value we will be using:
pOH = -log[0.01]
pOH = 2
Since the pOH is 2 and pH = 14 - poH
pH = 14 - 2
pH = 12, letter B