Final answer:
The pH of the solution prepared by dissolving 1.11 g of BaH2 in 777.1 mL of water is 13.01.
Step-by-step explanation:
To find the pH of the solution prepared by dissolving 1.11 g of BaH2 in 777.1 mL of water, we need to calculate the concentration of Ba²⁺ ions in the solution. First, we need to calculate the number of moles of BaH2 by dividing its mass by its molar mass (1.11 g / 139.346 g/mol = 0.00797 mol). Since BaH2 dissociates into 1 Ba²⁺ ion in the solution, the molarity of Ba²⁺ is 0.00797 mol / 0.7771 L = 0.0103 M. To find the pH, we use the formula pH = -log[H+], where [H+] is the concentration of H+ ions. However, since BaH2 is a base, the concentration of OH- ions needs to be considered. By using the equation Kw = [H+][OH-], where Kw is the ion product of water (1.0 x 10^-14), we can calculate the concentration of OH- ions as 1.0 x 10^-14 / 0.0103 = 9.71 x 10^-13 M. Using the equation pH = 14 - pOH, we get pH = 14 - (-log(9.71 x 10^-13)) = 13.01.