Question

If 2.73 g KHC2O4 H2C2O4 (three ionizable protons) having 2.0% inert impurities and 1.68 g KHC8H4O4 (one ionizable proton) are dissolved in water and diluted to 250 mL, what are the molarity and normality of the solution as an acid, assuming complete ionization?

Answer 1

Given the mass of potassium hydrogen phthalate(KHC8H4O4) in the sample = 1.68 g

molecular mass of potassium hydrogen phthalate(KHC8H4O4) = 204.22 g/mol

Hence moles of potassium hydrogen phthalate(KHC8H4O4) =  1.68 g / 204.22 g/mol = 0.00823 mol

mass of impurities = (2/100) x 2.73 g = 0.0546 g

mass of potassium tetraoxalate(KHC2O4 H2C2O4) = 2.73 g - 1.68 g - 0.0546 g = 0.9954 g

molecular mass of potassium tetraoxalate(KHC2O4 H2C2O4) = 218.16 g/mol

Hence moles of tetraoxalate(KHC2O4 H2C2O4) = 0.9954 g / 218.16 g/mol = 0.00456 mol

Hence total moles of both the acid in the solution = 0.00823 mol + 0.00456 mol = 0.01279 mol

Volume of the solution = 250 mL = 0.250 L

Hence molariy of the solution = moles of acid / V(L) = 0.01279 mol / 0.250L = 0.0512 M (answer)

Total number of replacible H - atoms = 3+1 = 4

Hence normality = 4 x molarity = 4 x 0.0512 M = 0.205 N (answer)