Experiments show that the solubility of alum in 25 mL of 1.4 M KOH plus 10 mL of

Question

Experiments show that the solubility of alum in 25 mL of 1.4 M KOH plus 10 mL of 9 M H2SO4 is about 1.0 g @ 1.0 degree Celsius and 1.7 g @ 6.0 degrees Celsius. Using your measured solution temperature, estimate the amount of alum left in your chilled solution.

Alum left in solution = ________g

Temp. cooled alum crystals = 5.3 degrees Celsius

I think you need to do a linear regression to figure out the alum dissolved at 5 degrees C.

(Change in Solubility)/(Change in temp) = (1.7 - 1.0)/(6 -1) = 0.7/5.3 = 0.1321

The solubility increases by 0.14 grams for each degree C above 1C.

At 5 degrees C the solubility will be 1 + (0.1321 x 5.3) = 1.70g

I just want to know if I did that right, or if I have to do  1 + (0.1321 x 4) = 1.53g, and if yes why do you multiply by 4 and not the 5.3?

Explanation / Answer

From your question, it seems that you need to calculate the solubility at 5 degree celsius. If that is the case then you need to multpily by 4. This is because you yourself have calculated that the solubility increases by 0.1321 grams for each degree C above 1C. Now the solubility at 1 degree C = 1.g. Now when you calculate solubility at 5 degree C, your increase in solubility = increase in solubility per degree C * increase in temp = 0.1321* (5 - 1) = 0.1321*4

But, this is only the INCREASE in solubility. To calculate the actual solubility at 5 degree C, you need to add this increase to the solubility at 1 degree C.

So, solubility at 5 degree C = solubility at 1 degree C + increase in solubility = 1 + (0.1321 x 4) = 1.53

Hope this clears your doubt. If not, you can comment below....!

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