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Determining Saturation


Silencer

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Is there a way that you can tell how concentrated a solution must be for it to be saturated?

 

For example, suppose I want to make a saturated NaCl solution in water. How do I know the maximum molarity possible and thus how much NaCl to use?

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But if you wanted to know the exact mass of salt needed for a solution to be saturated, you'd have to use the Ksp value. But if it doesn't matter what the mass is, then just keep adding salt until it doesn't dissolve. :D

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As much as i know therre is no easy way to compute solubilities. You have to make your own experiments or look for manuals. If anyone knows that those data can be computed someway then i'd like to know more about it. It would be very usefull - so i get rid of some heavy books on my table.

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Well, the Ksp of NaCl at 25 degrees Celcius is 37.3. So the equation is Ksp (37.3) = [Na+][Cl-]. Since one mole of NaCl results in a mole of Na+ and Cl- upon dissolving, you can say that 37.3 = x^2 where x = concentration of the salt at saturation. So the square root of 37.3 = 6.11. Therefore, the concentration of NaCl in a saturated solution is 6.11 moles/L, or 58.44 grams per Liter at 25 degrees Celcius.

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This can be done for nearly every Ksp value. Since the Ksp is an equillibrium constant at saturation for a dissolved salt. Since the Ksp is calculated from these concentrations at saturation, the concentrations can be determined. If you have a salt that's not binary, it's still easy to calculate. Let's take Na2SO4 for example. (Sadly, I cannot find the Ksp of this salt. It's a pain in the ass if you don't have a CRC Handbook to find Ksp values of soluble salts). So we'll assume the Ksp of Na2SO4 is 40.8. We would write the equation as 40.8 = [Na+]^2[sO4(2-)]. We'll replace the concentration of SO4(2-) with 'x', and the concentration of Na+ with '2x' since the concentration of sodium ions will always be twice that of sulfate ions. We now have 40.8=((2x)^2)(x) = 4x^3 -> 10.2 = x^3. x = concentration of the salt at saturation = 2.17 moles/L.

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