To 0.360 l of 0.160 m nh3 is added 0.120 m mgcl2. how many grams of (nh4)2so4 should be present to prevent the precipitation of mg(oh)2? the ksp of mg(oh)2 is 1.8x10-11.

first calculate the moles of every compound used

n NH3 = (0.160 mol /L )(0 .360 L) = 0.058 moles NH3
which is equal to moles NH4OH
0.058 moles NH4OH.

  n MgCl2 = (0.12 mol /L)( 0.10 liters = .012 moles MgCl2

reaction 2 NH3.H20 or 2 NH4OH + MgCl2 --> Mg(OH)2(s) + 2 NH4+ + 2Cl-

Molarity of Mg++ =(0.012 moles Mg+) / 0.46 liters = 0.026 Molar Mg++

Molarity OH= (0.058 moles OH-)/ 0.46 liters = 0.126 molar OH-

the limiting reactant is Mg ++ because it is lesser than the molarity of OH-

Now the challenge is to drive the OH-) concentration down so low that Mg(OH)2 will not precipitate out.

Ksp = 1.8 x 10^-11 = (Mg)(OH-)^2

  Mg++ concentration to be .026 Molar, so let X = the (OH-) concentration

1.8 x l0^-11 = (0.026)(X)^2

(X)^2 = ( 1.8 x 10^-11) / (0.026)

X^2 = 6.92 x 10^-10
X = 2.63 x l0^-5 moles (OH-)/ L
this is the concentration where solid will form

so we need to lower the (OH-) which is 0.126 molar OH- down to 2.63 x 10^-5 molar OH- by adding NH4+ ions.

(0.126 moles/liter 0H-) - (2.63 x l0^-5 molar OH- ) = 0.123 moles per liter

OH- to be neutralized by adding NH4+

since the mole ratio is 1 : 1 then  0.123 moles per liter NH4+ concentration to neutralize the  0.123 moles of OH- in solution.

so to prevent the precipitation of mg(oh)2

0.123 - 0.058 = 0.065 Molar NH4+ is needed