Question

A standard lanthanum solution is prepared by dissolving 0.1968 grams of lanthanum oxide (La₂O₃) in excess nitric acid and diluting to one liter in a volumetric flask. Calculate the concentration of the solution expressed as molarity of lanthanum nitrate and as ppm lanthanum

Answer 1

mass of lanthanum oxide = 0.1968 g

molar mass of La2O3 = 325.81 g/mol

moles = 0.1968 / 325.81 = 6.04 x 10^-4 moles

volume = 1 L

molarity = moles / volume = 6.04 x 10^-4 / 1 = 6.04 x 10^-4 M

molarity = 6.04 x 10^-4 M

La2O3 + 6 HNO3   ------------------> 2 La(NO3)3 + 3 H2O

for 1 mol of La2O3 gives -------------- 2 mol of La(NO3)3

6.04 x 10^-4 mol La2O3   --------------- ??

moles of La(NO3)3 = 2 x 6.04 x 10^-4

                                 = 1.208 x 10^-3

Molarity of La(NO3)3 = 1.208 x 10^-3 / 1

                                   = 1.208 x 10^-3 M

Molarity of La(NO3)3 = 1.208 x 10^-3 M

concentration of La = 1.208 x 10^-3 mol/L

                                = 1.208 x 10^-3 x 138.9

                                = 0.1678 g/L

                                = 167.8 mg / L

concentration of La = 167.8 ppm