Question

A solution is prepared by dissolving 7.82 g NaOH and 9.26 g Ba(OH)2 in water and diluting to 500. mL. What is the normality of the solution as a base

Answer 1

mass NaOH = 7.82 g

moles NaOH = (mass NaOH) / (molar mass NaOH)

moles NaOH = (7.82 g) / (40.0 g/mol)

moles NaOH = 0.1955 mol

moles OH^- from NaOH = moles NaOH

moles OH^- from NaOH = 0.1955 mol

mass Ba(OH)₂ = 9.26 g

moles Ba(OH)₂ = (mass Ba(OH)₂) / (molar mass Ba(OH)₂)

moles Ba(OH)₂ = (9.26 g) / (171.34 g/mol)

moles Ba(OH)₂ = 0.0540 mol

moles OH^- from Ba(OH)₂ = 2 * (moles Ba(OH)₂)

moles OH^- from Ba(OH)₂ = 2 * (0.0540 mol)

moles OH^- from Ba(OH)₂ = 0.1081 mol

Total moles OH^- = (moles OH^- from NaOH) + (moles OH^- from Ba(OH)₂)

Total moles OH^- = (0.1955 mol) + (0.1081 mol)

Total moles OH^- = 0.3036 mol

equivalents OH^- = Total moles OH^-

equivalents OH^- = 0.3036 eq

Total volume = 500 mL = 0.500 L

normality OH^- = (equivalents OH^-) / (total volume in Liter)

normality OH^- = (0.3036 eq) / (0.500 L)

normality OH^- = 0.607 N

normality of solution as a base = 0.607 N