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Author Topic: Solubility at Chemical Equilibrium  (Read 552 times)

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platypus1234

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Solubility at Chemical Equilibrium
« on: September 12, 2018, 12:12:55 PM »

 Hello, I need help with a solubility problem, I am not looking for the answer just how I should go about solving.

Solubility product of CaCO3 at 25C is 2.5x10^-9

a) calculate the solubility of CaCo3 in pure water (mol/L and g/100mL)
b) Calculate the solubility of CaCo3 in a solution of 0.2M CaCl2 at 25C.
c) what is the ratio of SO42- to CO32- in a solution at equilibrium with CaSO4 and CaCO3?

For a) I got x= 6.7*10^-5 mol/L and 6.7*10^-4 g/100mL

 I did this by
ksp=[Ca2+][CO32-]=[ x][ x]=4.5*10^-9
x^2=4.5x10^-9

and b) I got 2.3*10^-8 mol/L

I did this one by 0.2x=4.5*10^-9

I'm not confident on either of those answers, I just roughly followed the professors powerpoint and even if I did a and b correctly I dont understand why and I am lost on how I should even attempt c) ???
Thank You
« Last Edit: September 14, 2018, 02:59:07 AM by Borek »
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Borek

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Re: Solubility at Chemical Equilibrium
« Reply #1 on: September 14, 2018, 03:05:07 AM »

So far so good (although in never hurts to explain what you mean by x and why).

For c you will need Ksp of calcium sulfate. If both salts are present, what ions does the solution contain? Is there any dependency between concentration of Ca2+ and concentrations of CO32- and SO42-?
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