# Calculating Molar Solubility of CaCO3: A Puzzling Problem

• Chemistry
• samy4408
In summary: Then set up the Ksp expression for CaCO3 and solve for x, the conc of Ca ion, to find the molar solubility of CaCO3.In summary, the conversation discusses the problem of calculating the molar solubility of calcium carbonate (CaCO3) in water and in a solution of sodium carbonate (NaCO3) with a concentration of 0.1M. The Ksp for CaCO3 is given as 5*10^-9 and it is known that NaCO3 is completely soluble. The molar solubility in water is found to be 7.07*10^-5M, but there is confusion about the solubility in the NaCO3
samy4408
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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 !

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

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?

Borek said:
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

samy4408 said:
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.

"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.

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

chemisttree

## 1. What is molar solubility?

Molar solubility refers to the maximum amount of a solute that can be dissolved in a solvent at a given temperature, usually expressed in moles per liter (mol/L).

## 2. Why is calculating the molar solubility of CaCO3 considered a "puzzling problem"?

Calculating the molar solubility of CaCO3 is considered a "puzzling problem" because it involves various factors such as temperature, pH, and the presence of other ions, which can greatly affect the solubility of CaCO3. Additionally, CaCO3 can exist in different forms, such as calcite and aragonite, which have different solubility values.

## 3. How is the molar solubility of CaCO3 typically calculated?

The molar solubility of CaCO3 is typically calculated using the solubility product constant (Ksp) and the concentrations of Ca2+ and CO32- ions in the solution. The equation used is Ksp = [Ca2+][CO32-], where Ksp is the solubility product constant, [Ca2+] is the concentration of Ca2+ ions, and [CO32-] is the concentration of CO32- ions.

## 4. What are some challenges in accurately calculating the molar solubility of CaCO3?

Some challenges in accurately calculating the molar solubility of CaCO3 include variations in temperature and pH, the presence of other ions that can form complexes with Ca2+ and CO32-, and the different forms of CaCO3 that can exist. Additionally, experimental errors and limitations in measuring equipment can also affect the accuracy of the calculation.

## 5. How can the molar solubility of CaCO3 be used in practical applications?

The molar solubility of CaCO3 is important in various practical applications, such as in the production of cement and lime, water treatment, and in the pharmaceutical industry. It can also be used to determine the effectiveness of different compounds in inhibiting the precipitation of CaCO3, which is a common problem in many industrial processes.

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