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The solubility of calcium hydroxide

Topic: Chemistry
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The solubility of calcium hydroxide in water was determined. We would themselves come up with some solutions to the problem. The following methods were chosen:
First A saturated solution of calcium hydroxide are made and on the measured pH.
2nd A saturated solution of calcium hydroxide is titrated with hydrochloric acid predetermined concentration.
3rd A precise measured amount of saturated solution of calcium hydroxide dried until all the water is gone. This is then weighed.


Ca (OH) 2 + H 2 O ↔ Ca 2 + +2 OH-+ H2O
The solubility of a salt can be specified in two ways, the first in mass / volume (g/100 cm3), the other in solubility designated Ks. Ks is defined as the equilibrium constant, but the left side, in this case the Ca (OH) 2, is in solid form and therefore it has been determined that the concentration is 1. This means that


The saturated solution of calcium hydroxide was made by an excess of calcium hydroxide was added in 500 cm3 of water. The mixture was now turbid precipitates of calcium hydroxide was filtered through a filter paper made that the excess filtered off and only the dissolved calcium hydroxide ranns in solution. For this, we measured the pH with a pH meter.
The same solution was titrated now saturated with hydrochloric acid at a concentration of 0.02 M.
25.0 cm3 was measured with a pipette and placed in one, accurately weighed petri dish, this was set in a stove to dry. The solid salt in the bowl could now be considered.


Results from experiments

pH Ca (OH) 2 12.1225
V HCl 30.25 cm3
the pH at the equivalence point 6.765
Petri dish weight 6.7311 g
Petri weight of salt 6.7305 g

pH measurement
pOH = 14 to 12.1225 = 1.8775 => COH-10-1.8775 M = 0.0133  M =>

C (Ca (OH) 2 = 0.0133 / 2 = 0.0066, M = 0.00066 cm3 mol/100
M (Ca (OH) 2) = 74.026 g / mol
0.00066 mol/100cm3  74.026 g / mol  0.05 g/100cm3

C (HCl) = 0.02 M
V (HCl) = 0.03025 dm3
n (HCl) = 6.05 10-4 mol 

The molar ratio:
n (Ca (OH) 2) = 3.025  10-4 mol
V (Ca (OH) 2) = 0.025 dm3
C (Ca (OH) 2) = 0.0121 M
n (Ca2 +) = 3.025  10-4 mol
C (Ca2 +) = 0.0121 M
n (OH) = 6.05  10-4 mol
C (OH) = (6.05  10-4 mol) / 0.025 = 0.0242 dm3 M

n (Ca (OH) 2) = 3.025  10-4 mol
M (Ca (OH) 2) = 74.026 g / mol
m (Ca (OH) 2) = 0.022 g
V (Ca (OH) 2) = 0.025 dm3 = 0.25 (100cm3)
0.022 g / 0.25 (100 cm3) = 0.09 g/100cm3


One of the results can be discussed without the use of literature values. How can the petri dish without salt weigh more than salt? When we would take out the petri dish from the cabinet, where we let the water dry out, it was completely deformed. This should not have affected the weight of it, but apparently it happened. It may be that the bowl has become so hot that it has very little plastic dropping from it. This does not seem as likely as it would then move on so incredibly tiny droplets. Another option is that it would have incorporated gas of plastic, which actually sounds even more unbelievable. We did not have the heat on as high a temperature that evaporation would be likely. Instead, I would guess that it is we who have not been accurate enough when we have weighed the bowl, before and after. It may have been dirt that bothered weighing, or something else, but that's what I think is the reason for such an obviously incorrect result.
The solubility of calcium hydroxide is. Our values ​​and have at least the same order of magnitude as the true value, and they are so pretty good value even if you would like a little more of them. With the tools we used and including the human factor (at pH measurement), we should recognize the results approved!
Mass / unit volume of calcium hydroxide is 0.19 g/100 cm3. Our results, 0.09 and 0.05's then not so good value. This is probably due to the methods we chose. That we are not measuring lo sufficiently accurate, or that we got into some kind of error on the road. This would then also have affected the results Ks and they might have been better if we've made some error, but they may as well have been worse. This reasoning also requires results to be answered, it is not enough to bounce ideas back and forth.
Therefore, I am now leaving this and turns into something we can safely conclude. The choice of method to use is very important for how the results will be. The method gave the best results was the titration, which gave both the best results. The advantage of titration against only pH measurement is not so significant that the results should be so different. In fact, the pH measurement should give a more accurate result, as we did a pH measurement with a good pH meter. In the titration, we would determine exactly where the equivalence point was somewhere, which put the human factor, and the results are uncertain. It is thus more likely to have false results with the method that we got the best result. This indicates that an error has crept into the trials in one or another way. But as I wrote earlier, at least the results for Ks pretty good and therefore we should not blame the methods so much in this regard.


The solubility of calcium hydroxide was determined. This was done by using 1) evaporation, which was not so successful because the weight without salt was greater than the weight of salt.
Method 2) was that we were to measure the pH of a saturated solution of calcium hydroxide. This gave a Ks results were pretty much lower than what was actually required. Mass / volume unit value was 0.05, and literature values ​​is 0.18. our value was, therefore, bad.
The last method, Method 3) meant we titrated calcium hydroxide with hydrochloric acid. This method gave the best values.
What one can see from these experiments is that the method you choose is very important. You choose methods along the exact results you eat after, how good the instruments they have at their disposal and how long you want to put on trial. That our values ​​were so different was probably mainly due to the method.

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