Question

Determine the concentrations of MgCl2, Mg2+ and Cl- in a solution prepared by dissolving 1.94 x 10^-4 g MgCl2 in 1.25 L of water. Express all three concentrations in molarity. Additionally, express the concentrations of the ionic species in parts per million (ppm).

Answer 1

Solution :-

lets first calculate the moles of MgCl2

1.94*10^-4 g MgCl2 * 1 mol / 95.211 g = 2.04*10^-6 mol MgCl2

moles of Mg^2+ = 2.04*10^-6

moles of Cl- = 2.04*10^-6 * 2 = 4.08*10^-5

now lets calculate the concnetration of the MgCl2 in 1.25 L

molarity = moles / liter

              = 2.04*10^-6 mol / 1.25 L

              = 1.63*10^-6 M MgCl2

mole ratio of the Mg and MgCl2 is 1 : 1

so the molarity of the Mg^2+ = 1.63 *10^-6 M

mole ratio of the MgCl2 to Cl- is 1 : 2

so the concetration of the Cl- = 1.63*10^-6 * 2 = 3.26*10^-6 M

now lets calculate the mass of the Mg2+ and Cl-

mass = moles * molar mass

mass of Mg = 2.04*10^-6 * 24.305 g per mol = 4.958*10^-5 g   Mg2+

mass of Cl- = 4.08*10^-6 mol * 35.453 g per mol = 1.446*10^-4 g Cl-

ppm is mg per L

so

ppm of Mg^2+ = (4.958*10^-5 g * 1000 mg / 1 g) / 1.25 L = 0.0397 ppm

ppm of Cl- = (1.446*10^-4 g * 1000 mg / 1 g)/1.25 L = 0.1157 ppm Cl-