Question

Determine the concentrations of MgCl2, Mg2 , and Cl– in a solution prepared by dissolving 2.28 × 10–4 g MgCl2 in 1.75 L of water. Express all three concentrations in molarity. Additionally, express the concentrations of the ionic species in parts per thousand (ppt).

Answer 1

Determine the concentrations of MgCl2, Mg2 , and Cl– in a solution prepared by dissolving 2.28 × 10–4 g MgCl2 in 1.75 L of water. Express all three concentrations in molarity. Additionally, express the concentrations of the ionic species in parts per thousand (ppt)

Solution :-

Lets calculate the moles of the MgCl2

Moles = mass / molar mass

Moles of MgCl2 = 2.28*10^-4 g / 95.211 g per mol = 2.395*10^-6 mol

Now lets calculate the molarity of the MgCl2

Molarity of MgCl2 = moles / volume

                                   =2.395*10^-6 mol / 1.75 L

                                  = 1.37*10^-6   M

Mole ratio of the MgCl2 and Mg2+ is 1 : 1

So the concentration of Mg^2+ = 1.37*10^-6 M

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

So the concentration of the Cl- = 1.37*10^-6 M * 2 = 2.74*10^-6 M

Now lets calculate the concentration in ppt

Mass of Mg2+ = 2.395*10^-6 mol *24.305 g per mol = 5.821*10^-5 g * 1000 mg / 1 g = 0.05821 mg

(0.05821 mg *1 ppm / 1.75 L)*(0.1 ppt / 1 ppm ) = 3.326*10^-3 ppt

ppt of Cl- = (2.935*10^-6 mol *2 )*(35.453 g / 1 mol ) *(1ppm/1.75 L)*(0.1 ppt / 1 ppm) = 1.19*10^-2 ppt