Answer:
[Ba^2+]= 0.400M
[OH-]= 0.800M
Explanation:
Step 1: Data given
Molarity ofBa(OH)2 = 0.400 M
Ba(OH)2 is a strong electrolyte, so 100 % dissociation
Step 2: The equation
Ba(OH)2 → Ba^2+ + 2OH-
In 1 mol Ba(OH)2 we have 1 mol Ba^2+ and 2 moles OH-
Step 3: Calculate the concentration
[Ba^2+]= 1* 0.400 M = 0.400M
[OH-]= 2*0.400 M = 0.800M