The "Split and Mix" Solution Method

What is the "split and mix" solution method ?

This is a technique to grow large crystals step by step using the slowly cooling method. The goal is to get a warm solution which is almost perfectly saturated at this temperature. In this solution you can put smaller crystals or specimen to grow them bigger.

Lets give an example :

Take 1000 gms of a cold saturated solution of potassium alum, lets say its temperature is 20 C°, split in two portions of 500 gms each. Take one portion of 500 gms add some extra amount of potassium alum and heat it up until it boils (that is about 100 C°) then mix the cold and the hot solution to receive a warm solution which is exactly saturated at this resulting temperature.

There are two problems to solve :

1. What will be the resulting temperature ?

2. How much alum needs to be added to the boiling hot solution that the resulting solution is exactly saturated ?

The first problem is easy to solve if you use some simplifications. Just multiply the mass of the cold solution (mc) with its temperature (Tc) and the mass of the hot solution (mh) with its temperature (Th), add the two products and divide by the summ of the two masses combined and you get the temperature of the mixed solution (Tm).
As a mathematical formula it looks like that :

 (mc · Tc) + (mh · Th) Tm = (mc + mh)

In our example that will be with :

 mass of the cold solution (mc) = 500 gms temperature of the cold solution (Tc) = 20 C° mass of the hot solution (mh) = 500 gms temperature of the hot solution (Th) = 100 C°

 (500 gms · 20 C°) + (500 gms · 100 C°) Tm = = 60 C° (500 gms + 500 gms)

Okay, simplification one is of course that the total mass of our hot solution will be more than 500 gms as we are adding some extra amount of alum. This will result in a slightly higher temperature of the resulting mixed solution.
(However the water content of the solution carrys the major amount of heat energy not the disolved alum.)
Simplification two is that we do not pay any attention to the fact that mixing two solutions of different concentrations has an additional caloric effect (mostly this will result in a slightly lower temperature). Simplification one and two might compensate each other and the error in our calculation may be neglected.

Now we have a good clue what the temperature of our mixed solution will be.

Step two is to calculate the amount of alum which has to be added to the boiling hot solution. We need to know the saturation concentration of the cold and of the mixed solution as the difference between them has to be added.

The saturation concentration or the solubility of a substance is given in a solubility table at different temperatures in gms per 100 gms of water. Here is the solubility table of potassium alum :

solubility of potassium alum in water
in gms per 100 gms of water

 0 10 20 30 40 50 60 70 80 C° 5.7 8.5 12 18.5 25 36.8 58.5 94.4 195 gms/100gms

As you can see the saturation concentration
at 20 C° is
12.0 gms alum/100 gms water and
58.5 gms alum/100 gms water
at 60 C°.

So the difference (that what you have to add) is 46.5 gms/100gms of water.

How much water is in the 1000 gms of cold solution we have been starting with ?
We know 112 gms of cold solution contain 100 gms of water and 12 gms of alum (as the solubility table says).

so the 1000 gms of total starting solution contain :

 1000 · 100 = 892.85 gms of water 112

in your boiling hot solution part of 500 gms you have to disolve an extra mount of :

 892.85 · 46.5 = 415.17 gms of alum 100

So your complete recipe is :

Take 1000 gms of cold saturated solution of potassium alum split in two portions of 500 gms each.
Heat up one portion until it boils and disolve an extra of 415.17 gms of potassium alum in it.
Mix the two solutions and let them cool down.

Oops ! One last check. Are you shure you will be able to disolve this additional 415.17 gms of alum in the 500 gms of cold saturated solution ?
Okay the 500 gms of cold saturated solution contain

 500 · 12 = 53.57 gms of alum 112

and

 500 · 100 = 446.43 gms of water 112

so with the 415.17 gms of alum added you result will be a concentration of

 53.57 + 415.17 · 100 = 105 gms of alum / 100 gms of water 446.43

Well as the solubility of alum in water at 80 C° is already 195 gms/100 gms of water that should be no problem !

However you might not hit the exact saturation concentration and temperature.
It may be just because we made some simplifications.
The solubility data may not be correct.
There may have been errors in weighing and measuring the amounts of solution and alum added.

If you receive an undersaturated solution by this mehod your crystals may disolve.
To take off 5 or 10 % from the calculated add-on amount has been helpfull to prevent this.

For better results you always have to adjust the add-on amount. Don´t stick to the calculated value it can only be a good estimate.

Using the slowly cooling method to grow specimens the add-on amount should be in the range of 20 - 60 gms per 100 gms of water. For single crystals you better work with an add-on amount of 5 - 30 gms per 100 gms of water.

However keep in mind, the only thing you can vary is the ratio of how to split cold and hot solution.
Calculating on the base of this ratio will give you the add-on amount.

If for example you choose to split into a large amount of cold and a small amount of hot solution, your mixed solution temperature will be lower, so also the saturation concentration and ergo the add-on amount needed.
The possible errors in this scenario are lower than the other way round with a large amount of hot solution and only a small amount of cold solution.

Now this example is a little bit constructed to keep it simple. But what happens if the temperature of your cold solution is lets say 17 C° or 24 C°?
What happens if the temperature for the mixed solution calculates to, may be 64 C° or 47 C°?
There is no solubility table so accurate. Well its possible with some mathematical tricks to create a function who gives solubilty data for any given temperature (interpolation polynom).
I did not yet convert my old interpolotion polynom programm which worked on an ATARI and did invert matrixes and other tricky stuff to draw a solubility chart and calculate inbetwen values.
For the heck of it I wrote a little JavaScript which can read in a solubility table with cut and paste and calculate an approximated function of the form y = e(a+bx).

To pop up the programm click button below !

However a simple linear interpolation is also useable.
(solubility tables are not always perfectly exact...)
You can also estimate inbetween solubility values from a solubility diagramm or chart.

With Excel I got this diagramm:
(as obvious Excel can´t do interpolation polynoms !)

To calculate an inbetween solubility value you need a temperature and solubility value below the temperature in question (which we can call Tx).
Lets call them lower border values with Tl (lower border temperature) and Sl (lower border solubility).
And a temperature and solubility value above the temperature in question which we call Th (higher border Temperature) and Sh (higher border solubility).
These lower and higher border values can be picked from a solubility table and should of course be as close to Tx as possible.
The inbetween solubility value Sx for an inbetween Temperature Tx calculates as follows :

 (Tx - Tl ) · (Sh - Sl ) Sx = Sl + (Th - Tl )

To give an example we want to calculate the solubility of potassium alum at 53 C°. As lower border values we choose 50 C° with a solubility of 36.8 g/100g and as higher border values 60 C° with a solubility of 58.5 g/100g.

 (53 - 50 ) · (58.5 - 36.8 ) Sx = 36.8 + = 43.31 g/100g (60 - 50)

To much formulas ? Did not get it ? To much calculating back and for ?

There is hope !

This whole calculations can be done with the java-scripted "split and mix" solution calculator !

Just click on the button above.