Achemist dissolves 274.mg of pure barium hydroxide in enough water to make up 50.ml of solution. calculate the ph of the solution. (the temperature of the solution is 25°c.)

12.81.

Explanation:

M = (no. of moles of solute)/(Volume of the solution (L))

M = (mass/molar mass) NaCl / (Volume of the solution (L))

mass of Ba(OH)₂ = 274 mg = 0.274 g, molar mass of Ba(OH)₂ = 171.34 g/mol, Volume of water = 50.0 mL = 0.05 L.

M = (mass/molar mass) Ba(OH)₂ / (Volume of the solution (L)) = (0.274 g / 171.34 g/mol) / (0.05 L) = 0.03 M.

Ba(OH)₂ → Ba²⁺ + 2OH⁻,

Every 1.0 mol of Ba(OH)₂ gives 2.0 moles of OH⁻.

∴ [OH⁻] = 2(0.032 M) = 0.064 M.

∵ pOH = -log[OH⁻]

∴ pOH = -log(0.064) = 1.194.

∵ pH + pOH = 14.

∴ pH = 14 - pOH = 14 - 1.194 = 12.81.

every chemical equation adheres to the law of conservation of mass, which states that matter cannot be created or destroyed. therefore, there must be the same number of atoms of each element on each side of a chemical equation.