I hope this can help you. Below are the data from International Critical Tables, v. IV, P. 242, McGraw-hill, 1928, New York. After you may perhaps fit a curve like

ln (x) = A+ B/T+ C lnT or other you choose. 

Good luck!

DATA:

KAl(SO4)2

S.12aq

temperature (°C)    M +/- 3%

0                                0.177

10                              0,162

20                              0,228

30                              0,320

40                              0,458

50                              0,664

60                              0,948

70                             1.390

80                             2.12

85                             2.72

90                             3.48

where: S means anhydrous substance

M= Gram Formula weight per 1000 g water and 

mol fraction=x= M/(M+55,51)

Thanks! I think that this is the same chart everyone uses. Funny its almost a 100 years old. Well I guess some things don't change. Like physical laws! Id be happy to turn the formula into a spreadsheet but I dont recognize these variables. Im guessing there is Temperature, mass of solute, mass of water and time but not sure what goes where.

?

ln (x) = A+ B/T+ C lnT or other you choose.

sorry, I forgot to put in the system: x is the mol fraction , that you can calculate by the given formula,  mol fraction=x= M/(M+55,51) The parameters A, B and C you get from a multilinear regression of the data points. To do that, input for each x and T from the table, one column with ln x, other with 1/T (in K) and other of ln T (also in K). perform a multilinerar regression analysis to get A, B and C. Is a good ideai to recalculate the experimental points to see the degree of ajustment.

Good luck.

Hi Silas! Im looking at the figures you gave me and Im wondering if there isnt an error somewhere. The chart about that I posted is in Grams per 100ml. at 60 degrees the solution would be soluble at approximately 40g/100 ml or 400g/liter.

Your chart would put that in at about .948 grams,- a difference of 400%.

Is there an error somewhere?

In this curve I found in the critical tables, you must multiply by the number by the gram formula weight 258,21 for potash aluminum sulfate. this way at 60C it will be:

0.948*258,21= 244,78 g/1000 g of water or 24,48g/100g of water. In fact it is closer to the value obtained from your figure: ~25 g/100 g of water and not 40. 

40g/100 g of water in your figure is a little bit bellow  70°C. In the table, for 70 °C the value is 1.39 that, multiplied by 258,21 (sorry I just taken the data exactly from the original) is 358,91/1000 g of water or 35,9 g/100g of water. In fact there is a difference in the values, you have to check some points.

I just tried to give you an information in a table form that is better to adjust curves.

Good luke! 

Hello, again. There is a solubility-temperature  table in Perry's Chemical Engeneering Handbook in the chapter of physico-chemical properties, where you also can find the data of aluminun potash sulfate. 

Good luck!

Thank you very much for your help, even if it has been a long time after your last answer. Your question and answers are very helpfull.