Final Answer:
The molar solubility 
 of the salt 
 in a solution containing 
 is determined by the expression 
 is the solubility product constant. Solving for 
 yields the molar solubility.
Step-by-step explanation:
To find the molar solubility 
 of the salt 
, we start with the equilibrium expression for the dissolution of the salt:
![\[ MX \rightleftharpoons M^(+)_x + xX^(-) \]](https://img.qammunity.org/2024/formulas/chemistry/high-school/y9k9i696mkckxqs0tlq3vxpzy7xpuoshem.png)
The equilibrium constant 
 for this reaction is expressed as 
. Since the stoichiometry is 1:1 for 

Now, the solution already contains 
 which contributes to the concentration of 
 ions. Thus, the equilibrium concentration of 
. Substituting this into the 

Solving this quadratic equation for 
 we obtain the molar solubility. It's important to note that this approach considers the common ion effect from the initial concentration of 
. Understanding these principles is fundamental in predicting the behavior of salts in solution.
Lastly, calculating 
 provides insights into the saturation point of the solution, helping to assess potential precipitation or the formation of complexes.
The complete question is:
"What is the molar solubility (in mol L⁻¹) of a salt with the general molecular formula 
 is a cation and 
 is an anion) in a solution that initially contains 
? Given the solubility product constant 
 determine the equilibrium concentration of 
 in the solution."