Question

You prepare a Mg²⁺ standard solution by dissolving 0.101 g of solid MgSO₄x7H₂O in a 100mL volumetric flask and then performing a dilution of this solution into a 50 mL volumetric flask so that the dilute stock solution is one-fifth the concentration of the concentrated stock solution. Finally, you can dilute this second “dilute” stock solution into 100 mL volumetric flasks to achieve the desired standard concentrations. 1. What will be the concentration of Mg2+ ions, in both molarity and ppm, in the 50 mL volumetric flask? 2. What volume of the dilute stock solution will you pipet into the final 100 mL volumetric flask to make your 0.600 ppm standard?

Answer 1

Solution :-

0.101 g solid MgSO4.7H2O

Molar mass of MgSO4(H2O)7 is 246.4746 g/mol

Lets calculate the mass of the Mg in this sample

0.101 g * 24.305 g / 246.4746 g = 0.00996 g Mg

Now lets calculate the moles of Mg2+

Moles of Mg2+ = 0.00996 g / 24.305 g per mol = 0.00041 moles

0.00041 mol Mg2+ is dissolved in 100 ml volumetric flask

So the concentration of the 100 ml solution is

0.00041 mol / 0.100 L = 0.0041 M

Then it is dissolved in 50 ml volumetric flask so that concetration of the diluted solution is 1/5 of the initial concentration

So the diluted solution concentration = 0.0041 M / 5 = 0.00082 M

So lets calculate the concentrated stock solution needed to make this 50 ml solution

M1 = 0.0041 M , V1 = ? , M2 = 0.00082 M , V2 = 50 ml

V1= M2V2/M1

    = 0.00082 M * 50 ml / 0.0041 M

    = 10 ml

So the volume of the stock solution needed is 10 ml

So the molarity of the 50 ml diluted solution is 0.00082 M

Lets calculate the ppm concentration of the this

100 ml original solution contains 0.00996 g Mg2+

So in 10 ml solution mass of Mg2+ present is 0.00996 g * 10 ml / 100 ml = 0.000996 g

Ppm is 1 mg per L

So

0.000996 g * 1000 mg / 1 g = 0.996 mg

Ppm = 0.996 mg / 0.050 L = 19.92 ppm

So the ppm concetration of the 50 ml solution is 19.92 ppm

Now lets calculate the volume of the diluted solution needed to make 0.600 ppm solution

C1 = 19.92 ppm , V1 = ?

C2 = 0.600 ppm , V2 = 100 ml

V1 = C2V2/C1

    = 0.600 ppm * 100 ml / 19.92 ppm

   = 3.012 ml

So we need to take out 3.012 ml of the 19.92 ppm solution and dilute it in 100 mll flask to get the 0.600 ppm solution

So the molarity of the 50 ml diluted solution is 0.00082 M

So the ppm concetration of the 50 ml solution is 19.92 ppm

So we need to take out 3.012 ml of the 19.92 ppm solution and dilute it in 100 mll flask to get the 0.600 ppm solution