If water is saturated with CaCO3, how much of it has to evaporate to deposit .250g of CaCO3.
Ksp CaCO3 = 4.96 x 10^-7
Ok so anyway I got CaCO3 <===> Ca + CO3
Ksp = 4.96x 10^-9 = x^2
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Sq root = 7.043x10^-5 I am lost on where togo from here IF I am even doing it right
There is a discrepancy in your Ksp. You have E-7 in the problem but you substituted E-9. Check to make sure that is ok.
When you solve for x, that gives you the solubility CaCO3 in moles/L. I obtained about 7E-5 M and that x molar mass CaCO3 = about 7E-3 or 0.007 g CaCO3/L
Then I realized 1 L evaporated would only give me 0.007 g which isn't enough. Therefore, I took 100 L which gives 0.7 g. Now, what volume will contain 0.25g? That will be 100 L x 0.25/0.7 = about 36 L (approximate--you need to do it more accurately). So if you evaporated 36 L of 100L to leave 64L you would ppt 250 mg CaCO3. Your problem didn't specify how much of the saturated solution you had to work with.
Thanks Dr Bob The Ksp = 4.96 x 10^-9, so the second one I was using was the correct one, Reaching the M wasn't the hard part But I am still not sure of the equation. I will ask my teacher today in class. Somehow I came up with 52.133L but I did not factor the H2O in there anywhere , not sure if I even had to. And If I had to they gave me no saturation level to work with.
Thanks again !!
To answer this question, we need to use the solubility product constant (Ksp) of CaCO3 and the given amount of CaCO3 to calculate the concentration of the dissolved CaCO3 in water.
Let's assume that x g of CaCO3 dissolves in the water. Since the water is saturated, the concentration of CaCO3 in the water is equal to the solubility of CaCO3, which we can represent as [CaCO3].
Using the formula for Ksp:
Ksp = [Ca2+][CO32-]
Since CaCO3 dissociates into one Ca2+ ion and one CO32- ion in water, the concentration of Ca2+ ions is equal to [CaCO3], and the concentration of CO32- ions is also equal to [CaCO3].
Therefore, we can write the Ksp expression as:
Ksp = [CaCO3]²
Substituting the known value,
4.96 x 10^-7 = ([CaCO3])²
Taking the square root of both sides,
[CaCO3] = sqrt(4.96 x 10^-7)
Now, we have the concentration of CaCO3 in the water.
To find the amount of water that needs to evaporate to deposit 0.250g of CaCO3, we can use this concentration.
First, calculate the number of moles of CaCO3:
moles of CaCO3 = mass / molar mass of CaCO3
The molar mass of CaCO3 is:
40.08 g/mol (Ca) + 12.01 g/mol (C) + (3 * 16.00 g/mol) (O) = 100.09 g/mol
moles of CaCO3 = 0.250 g / 100.09 g/mol
Next, we can use the concentration of CaCO3 to calculate the total volume of the water that contains this amount of CaCO3.
Concentration (mol/L) = moles / volume (L)
Rearranging the equation, we get:
Volume (L) = moles / concentration (mol/L)
Finally, subtract this volume from the original volume of water to find the volume of water that needs to evaporate.
Keep in mind that this is assuming the volume changes linearly with the amount of CaCO3 dissolved.