posted by John on .
A 500ml saturated solution of MgCO3 (M=84) is reduced to 120 ml by evaporation. What mass of solid MgCO3 is formed?
The answer is 0.20g, I just need a detailed explanation.
Do you have a solubility or a Ksp to use. There is a good bit of disagreement on the net about the correct value for Ksp for MgCO3.
Oh sorry the Ksp=4x10^(-5)
MgCO3 ==> Mg^+2 + CO3^-2
Ksp = (Mg^+2)(CO3^-2) - (x)(x)
x = sqrt(4E-5) = 0.00632 moles/L or 0.00632 x 84 = 0.531 g/L
In 500 mL there will be 1/2 that or .265 grams.
If we evaporate water until the volume is 120 mL, then 0.531 g/L becomes
0.531 g/L x (120 mL/1000 mL) = 0.0637 g.
We started with 0.265 g in 500
............less 0.0637 in 120 mL =
solid ppting.....0.201 g which rounds to 0.2 g to 1 significant figures (from the Ksp value).
Thanks! But im confused on why you would subtract the .0637g in 120mL from .265g in 500mL. Why don't you do (grams from 500mL-grams from 380mL)?
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