Question

One way to remove magnesium (Mg²⁺) from water is to precipitate it with a strong base such as sodium hydroxide (NaOH). Assuming the following unbalanced reaction describes the process. (a) Calculate the concentration of NaOH (in mg/L) needed to completely remove 20 mg/L of Mg²⁺.

(b) Calculate the mass (in lb) of solid Mg(OH)₂ that will be produced (and must be disposed of) if this reaction is performed in 250,000 gallons of water.

MgSO₄ + NaOH = Mg(OH)₂¯ + Na₂SO₄

Answer 1

MgSO4---> Mg+2 + SO4-2

MgSO4 +2NaOH----> Mg(OH)2 + Na2SO4

1 mole of MgSO4 requires 2 mole of NaOH

MgSO4----> Mg+2 + SO4-2

1 mole of MgSO4 is equal to 1 mole of Mg+2

moles of Mg+2= mass/Molecular weight, Molecular weight of Mg=24

Moles of Mg+2= (20/24) Mgmoles/L=0.833 mgmoles/L

mg moles of NaOH required= 2time that of MgSO4= 2*0.833= 1.666 moles of NaOH/L

Molecular weight of NaOH= 40

Mass of NaOH required per L= 1.666*40 mg/L= 66.67 mg/L

b) 1 Gallon =3.78 L

250000 Gallon= 250000*3.78 L=945000 L

L contains 0.833 moles of Mg+2 =0.833 moles of MgSO4

945000 L contains 0.833*945000=787185 moles of MgSO4

1 mole of MgSO4 produces 1 mole of Mg(OH)2

787185 moles of MgSO4 produces 787185 moles of Mg(OH)2

Molecular weight of Mg(OH)2= 58

Mass of Mg(OH)2 produced = 781785*58=45656730 gms of Mg(OH)2=45656.73 kg of Mg(OH)2

1 kg = 2.2 lb

45656.73 lb= 45656.73*2.2 lb= 100445 lb

45656730

1 mole of MgSO4 prodcues 1 mole of Mg+2

m