This is an incomplete question, here is a complete question.
Calculate the solubility of each of the following compounds in moles per liter. Ignore any acid-base properties.
CaCO₃, Ksp = 8.7 × 10⁻⁹
Answer : The solubility of CaCO₃ is,
![9.33* 10^(-5)mol/L](https://img.qammunity.org/2021/formulas/chemistry/college/j67lh7qfdpgrcsvmimr966riivaltnkakn.png)
Explanation :
As we know that CaCO₃ dissociates to give
ion and
ion.
The solubility equilibrium reaction will be:
![CaCO_3\rightleftharpoons Ca^(2+)+CO_3^(2-)](https://img.qammunity.org/2021/formulas/chemistry/college/zqzvu9ebbfgyy1023om5esoxoqhhgoe0r2.png)
The expression for solubility constant for this reaction will be,
![K_(sp)=[Ca^(2+)][CO_3^(2-)]](https://img.qammunity.org/2021/formulas/chemistry/college/xyd5s3ein7jqgamaxt1hn0c20i8rq5fuv0.png)
Let solubility of CaCO₃ be, 's'
![K_(sp)=(s)* (s)](https://img.qammunity.org/2021/formulas/chemistry/college/2i10cozkbhxxsfswelmwxddx06f1uie9tg.png)
![K_(sp)=s^2](https://img.qammunity.org/2021/formulas/chemistry/college/f541oml1epzrfrlevwtwezqmybkhss9yv6.png)
![8.7* 10^(-9)=s^2](https://img.qammunity.org/2021/formulas/chemistry/college/w86sohf4i4re658nne5xp0l64f4edwl3em.png)
![s=9.33* 10^(-5)mol/L](https://img.qammunity.org/2021/formulas/chemistry/college/1l2endhdd97sxsf0i2gindo52zw3cwh06l.png)
Therefore, the solubility of CaCO₃ is,
![9.33* 10^(-5)mol/L](https://img.qammunity.org/2021/formulas/chemistry/college/j67lh7qfdpgrcsvmimr966riivaltnkakn.png)