Hi,
Thanks. Now we have the essential information to explain why your science fair project did not work. And, I’ll try to explain one step at a time so you can understand the chemistry.
1. Here is the composition of your samples:
28 Grams or .36 M calcium carbonate (CaCO3)(mussel shells).
15 ounce or .44 L ocean water
pH 8.1 or pH 7.5
http://www.metric-conversions.org/volum ... rs.htm?arg
2. Here is what happened during your experiment; this is the quantity of calcium carbonate that could have dissolved during the month at 25 degrees Centigrade in pure water.
During the month that the shells were immersed in the ocean water, some molecules of calcium carbonate would dissolve and come to equilibrium with the solid calcium carbonate. The solubility constant (Ksp) of calcium carbonate in the website I found today is given as 8.7 x 10-9, so we’ll use this number for the calculations.
http://en.wikipedia.org/wiki/Calcium_carbonate
The solubility constant of a compound is the maximum concentration of ions that can dissolve:
Ksp calcium carbonate = [Ca+2] [CO3-2] = 8.7 x 10-9
So the concentration of calcium and carbonate ions that could dissolve in water would be 9.3 x 10-5 M (moles per liter) of both calcium and carbonate ions.
http://www.math.com/students/calculator ... e-root.htm
Now, adjusting for the volume of your water, the amount of calcium and carbonate in your sample would be:
9.3 x 10-5 M/ L * .44 L = 4.2 x 10 -5 moles per jar
And converting this to grams
4.2 x 10 -5 moles/liter * 100 grams/mole = 4.2 x 10 -3 grams, or .0042 grams
So, you should have been able to detect a loss of 0.0042 grams of calcium carbonate during the month, if you had used pure water. If your scale is not accurate to the thousands of a gram level, then this could explain why you didn’t see any change in the weight of the mussel shells. I think that the scale your teacher has doesn’t have the accuracy required for this experiment either, so this is the primary reason for lack of results.
2. Effect of the pH difference.
Here is an explanation of the difference in the concentration of hydrogen ions between pH 7.5 and pH 8.1. You should include this explanation in your write up since you were doing a pH experiment and you will want the science fair judges to know that you understand pH.
http://en.wikipedia.org/wiki/PH
Here is the formula for calculating pH, which is the negative log of the hydrogen ion concentration:
Remember from yesterday’s post, 2 hydrogen ions are required to dissolve one molecule of calcium carbonate.
CaCO3(s) + 2 HCl(aq) → CaCl2(aq) + CO2(g) + H2O
The concentrations of hydrogen ions in your two samples were:
pH 8.1 = 7.9432823472e-9 M=concentration of hydrogen ions
pH 7.5 = 3.1622776602e- 8M= concentration of hydrogen ions
http://www.1728.com/logrithm.htm
Subtracting these numbers, you will see that you had 2.37 x 10 -8 more moles/L of hydrogen in the pH 7.5 sample compared to the pH 8.1 control. Adjusting for your volume (.44 liters), you would have had 1.04 x 10 -8 more moles in the pH 7.5 jar, so this would potentially dissolve 0.5 x 10 -9 more moles of calcium compared to the pH 8.1 jar. Converting this to weight, if every extra hydrogen ion had been used to dissolve the calcium carbonate, you would have seen the following difference in weight between your two jars.
0.5 x 10 -9 x 10 g/M = 0.5 x 10-7 grams, or 0.0000005 more grams of calcium carbonate dissolved in the pH 7.5 jar compared to the pH 8.1 jar.
3. Effect of water matrix.
There’s one more consideration that you need to account for. You used instant ocean, which contains some salts, including about 400 ppm calcium. What effect would that have on your results?
http://www.aquacraft.net/w0014.html
400 ppm calcium is equal to 0.4 g Ca/liter x M Ca/40 g= 0.01 M (moles per liter)
Since there are already calcium ions in solution, this will limit the amount of calcium carbonate that can dissolve. Using the solubility constant again, you can find the concentration of carbonate that could dissolve in 0.44 liters of instant ocean is:
[Ca+2] [CO3-2] = 8.7 x 10-9 =[0.01][CO3]
Solving for concentration of carbonate:
[CO3] = 8.7 x 10-9 / 0.01 = 8.7 x 10-7 moles of carbonate.
Since only 8.7 x 10-7 moles of carbonate could dissolve from the shells, then only the same concentration of calcium could have also dissolved. Converting this to grams,
8.7 x 10-7 moles of calcium carbonate x 100 grams/mole = 8.7 x 10-5 grams of calcium carbonate or 0.000087 grams. So if you had a balance with an accuracy to the nearest 10,000th of a gram, you would have been able to measure a loss of weight in your control sample, and if you had a balance with an accuracy to the nearest 1,000,000th of a gram, you would have been able to detect a difference between the control and the experimental sample.
So, there is lots of science in this project after all. And, I’m sure you did get some results, but you just couldn’t measure them. Be sure to write up the complete project; you should go ahead and graph your results, even though it’s going to be a flat line or flat bar graph, depending on what you decide to use. If it appeals to you, you can use some humor to explain your results. The above information should be used in the discussion section, and you should include a final paragraph describing what you would do differently if you had time to repeat the experiment (e.g. put 25 grams of shells in a container the size of a swimming pool).
And, please do let me know if you have any questions on the chemistry. This is a study in solubility constants and pH, and you need to really understand what is happening so you can explain your results.
http://en.wikipedia.org/wiki/Solubility_equilibrium
Donna Hardy