Calculating Molar Solubility of CaCO3: A Puzzling Problem

[samy4408]

Homework Statement

a homework about solubility

Relevant Equations

no equation

hello , i am solving a problem about solubility and the solution seems weird to me , the problem is the following :
we are asked to calculate the molar solubility of (CaCO3) with (Ks = 5*10^-9) in water and in a solution of (NaCO3) with a concentration of 0,1M, knowing that (NaCO3) is totally soluble
i found that the molar solubility in water was s=7,07*10^-5M
the problem is in the second question , the solubility in a solution of (NaCO3) , because [CO3-]>>s
(CaCO3) should not dissolve at all .
that don't match the correction , can anyone tell me what is wrong ?
thanks !

 

[Bystander]

https://www.google.com/search?q=com...57j0i512l9.17408j0j7&sourceid=chrome&ie=UTF-8

 

[Borek]

There is no such thing as "should not dissolve at all". It always dissolves till the Ksp is satisfied.

 

[Mayhem]

7.07E-5 M is a very low concentration. That's 0.0000707 M !
Are you familiar with ICE tables and how you use them to solve these kinds of problems?

 

[samy4408]

There is no such thing as "should not dissolve at all". It always dissolves till the Ksp is satisfied.

take this dissolution reaction :
AB => A+ + B-
if we have a large concentration of B- in the water such that [B-] >> Ksp , the dissolution of each molecule of AB will increase the concentration of [B-] (which means that we are going to move away from the equilibrium)
, and we don't have free A+ in the solution so i don't understand how the reaction can shift toward the left

 

[Borek]

take this dissolution reaction :
AB => A+ + B-
if we have a large concentration of B- in the water such that [B-] >> Ksp , the dissolution of each molecule of AB will increase the concentration of [B-] (which means that we are going to move away from the equilibrium)
, and we don't have free A+ in the solution so i don't understand how the reaction can shift toward the left

Molar solubility will be the concentration of A^+, so it is just

$$[A^+]=\frac{K_{sp}}{[B^-]}$$

You need to take into account all sources of B^- though, as @Mayhem already suggested ICE table is your friend here.

 

[epenguin]

"we are asked to calculate the molar solubility of (CaCO3) with (Ks = 5*10^-9) in water and in a solution of (NaCO3) with a concentration of 0,1M, knowing that (NaCO3) is totally soluble
i found that the molar solubility in water was s=7,07*10^-5M
the problem is in the second question , the solubility in a solution of (NaCO3) , because [CO3-]>>s"

Don't know whether we can help you. At least get the formulae in the question right.

 

[Chestermiller]

Let x be the Ca ion conc. Then the carbonate ion conc will be 0.1+x