Answer:
The pH of the solution formed is approximately 12.614.
Step-by-step explanation:
First, we need to calculate the molarity of the Ba(OH)₂ solution:
moles of Ba(OH)₂ = mass / molar mass = 3.470 g / (137.327 g/mol + 2*15.9994 g/mol) = 0.01156 mol
molarity = moles / volume = 0.01156 mol / 0.5631 L = 0.02055 M
Ba(OH)₂ is a strong base, and it will dissociate completely in water to give two OH⁻ ions for every Ba(OH)₂ molecule:
Ba(OH)₂ → Ba²⁺ + 2OH⁻
So the concentration of OH⁻ ions in the solution is 2 * 0.02055 M = 0.0411 M.
Now we can use the equation for the ion product constant of water to calculate the pH of the solution:
Kw = [H⁺][OH⁻] = 1.0 x 10⁻¹⁴
pH + pOH = 14.00
pOH = -log[OH⁻] = -log(0.0411) = 1.386
pH = 14.00 - pOH = 14.00 - 1.386 = 12.614
Therefore, the pH of the solution formed is approximately 12.614.