posted by maureen .
A 600ml sample of a saturated solution of MgCO3 is reduced to 125 ml by evaporation. What mass of MgCo3 is formed? Ksp=4*10^-5
MgCO3 ==>Mg^2+ + CO3^2-
Ksp = (Mg^2+)(CO3^2-)
If you let x = solublity MgCO3, then x = (Mg^&2+) and x = (CO3^2-). Substitute and solve for x = solubility MgCO3.
That will be mols/L. Convert to mols in 600 mL and call this y1 mols to start.
Then convert to mols in 125 mL and call this y2 mols. The difference between y2 and y1 will be what ppts. Convert that to grams.