Answer to Question #331341 in Chemistry for Jan

Question

If 24.15 ml of sodium hydroxide were required in the titration of a 4.9651 g sample of dried reagent potassium biphthalate (KHC8H4O4), what would be the normality of sodium hydroxide?

Answer 1

NaOH + KHC₈H₄O₄ = NaKC₈H₄O₄ + H₂O

n = m / M

M (KHC₈H₄O₄) = 204.22 g/mol

n (KHC₈H₄O₄) = 4.9651 / 204.22 = 0.024 mol

n (NaOH) = n (KHC₈H₄O₄) = 0.024 mol

C_M (NaOH) x V (NaOH) = n (KHC₈H₄O₄)

C_M (NaOH) = n (KHC₈H₄O₄) / V (NaOH) = 0.024 / 0.02415 = 1 M

When doing normal/molar conversions, remember when the valence is one, normality and molarity are the same. The normality of the 1 M solution of NaOH is 1 N.