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.
Thanks!
Question ID
524132
Created
April 2, 2011 2:33am UTC
Rating
0
URL
https://questions.llc/questions/524132
Answers
6
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6 answers
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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.
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Oh sorry the Ksp=4x10^(-5)
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OK. Thanks.
MgCO3 ==> Mg^+2 + CO3^-2
..x........x.......x
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)?
Answer ID
524361Created
April 2, 2011 10:14pm UTCRating
0 -
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1087723Created
August 26, 2014 9:19pm UTCRating
-1 -
Well, let's break this down step by step.
First, we need to find out the initial mass of MgCO3 present in the 500 ml saturated solution. To do this, we can use the formula:
Mass = (Concentration * Volume) / Molar Mass
The concentration of a saturated solution is the maximum amount of solute that can dissolve in a given amount of solvent at a given temperature. In this case, the concentration of MgCO3 is at its maximum, so we can use its solubility. The solubility of MgCO3 is 0.067 g/ml.
So, the initial mass of MgCO3 in the 500 ml solution is:
Mass = (0.067 g/ml * 500 ml) / 84 g/mol
Mass = 0.399 g
Now, we know that the solution was reduced to 120 ml by evaporation. We need to find the mass of the solid formed during this process. To do this, we can calculate the mass of the remaining solution and subtract it from the initial mass.
The final mass of the remaining solution is:
Final Mass = (0.067 g/ml * 120 ml) / 84 g/mol
Final Mass = 0.096 g
Now, to find the mass of solid MgCO3 formed, we subtract the final mass of the solution from the initial mass:
Mass of Solid = Initial Mass - Final Mass
Mass of Solid = 0.399 g - 0.096 g
Mass of Solid = 0.303 g
So, the mass of solid MgCO3 formed is 0.303 g. I'm sorry if this explanation was a bit dry, but I hope you found it useful!Answer ID
3062112Created
September 28, 2023 2:28pm UTCRating
0