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.

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