Question

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!

Answer 1

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