200.0 mL of a solution of 2.000M MgCl2 was mixed with equal amount of 2.000M Na2CO3and MgCO3 precipitate was collected (MgCO3 Ksp = 1.580x10?8). How much (how many grams) of the precipitate was collected? What is the final concentration of the [Mg] in the solution?
Steps would be appreciated!
V1 = 200 ml
M1 = 2M MgCl2
V2 = 200 ml
M2 = 2M Na2CO3
MgCO3 precipitate
Ksp = 1.58*10^-8
MgCO3 <-> Mg+2 and CO3-2
KSp = S*S
S^2 = 1.58*10^-8
s = sqrt(1.58*10^-8) = 0.00012569 M is soluble
[Mg+2] = 0.00012569 M is in solution
Since
Vt = 200 ml + 200 ml = 400 ml
M1*V1 = 2*0.2 = 0.4 mmol of Mg+2
M2*V2 = 2*.02 =0.4 mmol of CO3-2
0.4 mmol of MgCO3
M = mol/V = 0.4 mmol / 400 ml = 0.001 M of MgCO3 will be formed (in solution and non solution)
Since 0.00012569 mol per liter will be soluble..
M3 = mol3/Vt
0.000125639 = mol3 / 400 ml
mol3 = 0.000125639*0.4 = 0.0000502 mol of MgCO3 are in solution
0.001 mol are formed - 0.0000502 that are in solution = 0.0009498 mol are precipitated
MW = MgCO3
mass = mol*MW = 0.0009498 mol * 84.3 = 0.08 grams of MgCO3 will be collected
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