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I attempted to prepare saturated citric acid solution from anhydrous powder. According to online data, the solubility limit was 59.2 w/w at 20 °C, i.e. 59.2 g citric acid in 100 mL water. To ensure full saturation (as in the case of preparing brine solution), I ended up weighing 67 g of the powder and tried adding 100 mL of water to it.

However, when I had added about 67 mL of water, the 100 mL Schott bottle was already full, and all of the powder dissolved (estimated 5.22 M solution). Since there is an apparent volume expansion, how then should this solution (or a citric acid solution of any other desired concentration) be prepared?

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    $\begingroup$ Add some water till the solid stuff dissolves, then gradually add more and more until you hit 100ml. $\endgroup$ Feb 15 at 9:09
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    $\begingroup$ Concentrations are always given in [amount] per volume of solution, NOT per volume of solvent. $\endgroup$
    – Karl
    Feb 15 at 9:24
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The "expanding solution" problem is irrelevant if you use standard techniques

First, you need to realise that a standard solution is a known amount of solid in a known volume of solution not a known amount of solid and a known amount of the solvent. You don't need to know how much solvent you use. So you need to target getting a result that occupies, say, 100mL not a solution made from 100mL of water.

There is an easy way to achieve this. Use a volumetric flask with an accurate volume marking. Add the solid to the flask. Fill the flask slowly (while shaking enough to ensure the solid dissolves before the flask is full). Keep adding small increments of the solvent until the total volume reaches the graduation. You now have an exact volume of a solution containing an exact amount of the solid. At no point do you need to worry about any volume expansion or contraction.

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  • $\begingroup$ That is directly applicable with the goal to reach the given mass or molar concentration m/V or n/V. But not so directly if data are provided as solubility, mass fraction or molality. You need to know the target solution density. If not tabelized, you need to prepare the solution first and measure it. $\endgroup$
    – Poutnik
    Feb 26 at 11:20
  • $\begingroup$ Was the OP trying to prepare a "standard solution"? The expanded question asked for a solution of "any other desired concentration", which seems more like a mass to mass ratio, not mass to volume. $\endgroup$ Feb 26 at 14:30
  • $\begingroup$ Darn it! I just realized why there are two paths recommended for the calculations! Some of us chemists are thinking in moles per liter, while some of us (well, me) are (is) thinking in terms of %. In commercial practice, where we deal with hundreds or thousands of gallons, % is the way to go, even if you do the preparation on a lab scale. Academics might prefer volumetric flasks. I always had to standardize the flasks by weight - they never seemed accurate enough, or convenient enough, because I always had to add it to more stuff or more stuff to it. $\endgroup$ Feb 26 at 14:41
  • $\begingroup$ @JamesGaidis In bulk commercial preparation, you usually have a well-tested recipe specifying the amounts to mix in convenient units and whoever developed the recipe will have checked that the recipe works. In a lab you usually want accuracy (in g/L or something like that). Hence volumetric flasks and weighed solids. $\endgroup$
    – matt_black
    Feb 26 at 15:25
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This is an example of a Schott bottle:

enter image description here

Using the volume markings would not be very accurate, so you would weigh the water.

Citric acid solution is indeed less dense than you would calculate from the volumes and densities of water plus citric acid. 30% solution has a density of 1.13, but the arithmetically calculated density is 1.1995 from citric acid density = 1.665 and water = 1 (CRC Handbook, 62 ed). There were no data for more concentrated solutions.

As Poutnik points out, 59.2% w/w means 59.2% of the final solution is citric acid - yes, it's that soluble! But a better way to make the solution and fill the bottle to the 100 mL mark (if that is important) is to prepare the solution in another flask, by weight. Since you want 100 mL, which will weigh 113 grams, I suggest you go for 125 grams, to have a little extra. 125 x 0.592 = 74 grams citric acid; add 51 grams of warm water (to insure more rapid and complete solution) and swirl till clear. (Swirl the flask, not yourself.)

When the solution is clear, fill the Schott bottle. What to do with the excess solution? Do not fret. Citric acid solution is available commercially as a toilet bowl cleaner. 12 grams of saturated citric acid will not do much cleaning for a really dirty toilet bowl, but at least it won't damage the ecology any more than we normally do.

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One thing is usual solubility in g/100 mL, other thing is solubility as percentual mass fraction.

(Improperly used) 59.2 w/w obviously means 59.2 weight percent, as this solubility data – noted properly – is listed in wikipedia: Citric acid.

You need 59.2 g of the acid and 40.8 g of water ( about 40.8 mL or approx. 41 mL ), in total 100 g. The volume will be less than 100 mL, due the higher than water density.

Recalculated to solubility it is approximately $\pu{145 g/100 mL}$

In the case an acid monohydrate is used instead, the stoichiometric mass of crystal water has to be obviously subtracted from the acid mass and added to the water mass.

That can be recalculated for any target total mass and particular volume can be taken from this stock solution.


comments to comments:

@matt_black Sure, there are multiple ways to prepare saturated solutions.

  • One is the preparation of arbitrary mass of w/w% solution according to provided data.
  • Other is recalculation of w/w% to g/100 mL and dissolving in this nominal volume.
  • Other is reaching equilibrium of stirred solution with an excess of acid.
  • Other is to obtain density of saturated solution, calculating the needed acid mass as percentage of the total volume mass. And then not fully filling up into given volume. After near full dissolution, mixing/filling up in few steps until the nominal volume.
  • Other is to calculate the acid mass equivalent to the acid volume with the filling volume as a complement. But this is not recommended, being tricky, not applicable to well soluble and quickly dissolving compounds. As it implies filling is done before significant dissolution occurs, because there is volume change during dissolution.
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    $\begingroup$ Actually, nonconforming expressions such as "w/w" should not be used at all. $\endgroup$
    – Loong
    Feb 25 at 14:47
  • $\begingroup$ @Loong I do not object, it was used by the OP. Unless you mean in MathJax. I was in doubt, if I should use it. It could be plain text as well. $\endgroup$
    – Poutnik
    Feb 25 at 15:19
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    $\begingroup$ I am aware that "w/w" was used in the question; no need to change your answer. My comment was more a reflex reaction to your explanation "59.2 w/w means 59.2 weight percent" because actually such things don't mean anything since they must not be used. $\endgroup$
    – Loong
    Feb 25 at 15:32
  • $\begingroup$ You don't need to know the amount of water at all if you use standard techniques and volumetric flasks. $\endgroup$
    – matt_black
    Feb 26 at 9:59

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