Write a procedure you could use to prepare a 0.64 M solution of CaCO3 in the laboratory that would be ready for use
Write a procedure you could use to prepare a 0.64 M solution of CaCO3 in the laboratory that would be ready for use. Remember to be practical and specific (Scales measure grams not moles. Consider the type of measuring tools you would use, etc.).
`CaCO_3` is not fully soluble in water. It is a precipitate and dissolved partially in water.
`CaCO_3(s) hArr Ca^(2+)_(aq)+CO_3^(2-)_(aq)`
At 25 C the `K_(sp)` of CaCO_3 is between `3.7xx10^(-9) and 8.7xx10^(-9)` .
Let us say `K_(sp) = 6.4xx10^(-9) `
`K_(sp) = [Ca^(2+)][CO_3^(2-)]`
If the solubility of CaCO_3 at 25C is x M then;
`6.4xx10^(-9) = x^2`
`x = 8xx10^-5`
This means from 1 mole of `CaCO_3` we can get only `8xx10^-5` of ions for the `CaCO_3` solution.
The x value is well below 0.64M. So at room temperature we cannot prepare a `CaCO_3` solution of 0.64M.
When increased temperature the solubility will increase.
So first we have to find a temperature that solubility of `CaCO_3` is equal to 0.64M. At that temperature `K_(sp) = 0.64^2 = 4*10^-3`
If you find that temperature you can use the following method to prepare the sample.
The procedure is as follows.
- Take a flask which is capable of storing more than 1 liter of water
- Using a measuring cylinder measure 1 liter of clean water which is free of any impurities pour in to the flask.
- heat of the flask with water to the temperature which you found earlier and maintain the temperature. You may use a Bunsen burner for this task.
- Then take a pure sample (better if we can get in powder form) and weigh 100g (1mol) of with a balance which has a accuracy of 0.1g
- Then this weighted was put in to the flask carefully without misplacing them.
- Finally we can stir the water until `CaCO_3` dissolve and come to equilibrium.
- Now we have a 0.64M `CaCO_3` solution.
However my opinion is this is a useless task.