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

Answer:

1.208x10⁻³M and 392.5ppm La(NO3)3

Explanation:

The reaction that occurs is:

La2O3 + 6HNO3 → 2La(NO3)3 + 3H2O

Molarity is defined as the moles of solute (In this case, LaO3) per liter of solution. And ppm, are mg of solute per liter of solution.

To solve this question we must find the moles of La(NO3)3 produced and its mass in milligrams to find molarity and ppm:

Moles La2O3 -Molar mass: 325.81g/mol-

0.1968g * (1mol / 325.81g) = 6.04x10⁻⁴ moles La2O3

Moles La(NO3)3:

6.04x10⁻⁴ moles La2O3 * (2mol La(NO3)3 / 1mol La2O3) = 1.208x10⁻³ moles La(NO3)3

Molarity:

1.208x10⁻³ moles La(NO3)3 / 1L =

1.208x10⁻³M

Mass La(NO3)3 -Molar mass: 324.92g/mol-

1.208x10⁻³ moles La(NO3)3 * (324.92g / mol) = 0.392.5g La(NO3)3

In mg:

392.5mg La(NO3)3 / 1L =

392.5ppm La(NO3)3