Answer:
M of Ba(OH)₂ = 0.818 M.
Explanation:
Ba(OH)₂ + 2HNO₃ → Ba(NO₃)₂ + 2H₂O,
It is clear that every 1.0 mol of Ba(OH)₂ needs 2 mol of HNO₃ to be neutralized completely.
∴ (nMV) of Ba(OH)₂ = (nMV) for HNO₃.
where, n is the no. of producible H⁺ or OH⁻ of the acid or base, respectively.
M is the molarity of the acid or base.
V is the volume of the acid or base.
For Ba(OH)₂:
n = 2, M = ??? M, V = 20.0 mL.
For HNO₃:
n = 1, M = 0.85 M, V = 38.5 mL.
∴ M of Ba(OH)₂ = (nMV) of HNO₃ / (MV) for Ba(OH)₂ = (1)(0.85 M)(38.5 L)/(2)(20.0 mL) = 0.818 M.