Short Answer 
The solubility product constant (Ksp) of silver chloride (AgCl) is calculated as Ksp = [Ag‚A∫][Cl‚Aª], leading to a relationship where Ksp = x¬≤. Given Ksp = 1.8 √o 10‚Aª¬π‚A∞, the molar solubility is determined to be 1.3 √o 10‚Aª‚Aµ M, but with an increased chloride concentration of 0.54 M, the solubility decreases to 3.3 √o 10‚Aª¬π‚A∞ M due to the common ion effect.
Step 1: Understanding Solubility Product Constant
The solubility product constant (Ksp) of silver chloride (AgCl) can be calculated using the concentrations of silver ions (Ag+) and chloride ions (Cl–). Both ions are produced in equal amounts when AgCl dissolves, allowing us to set the equation as Ksp = [Ag+][Cl–]. If we let the concentration of both ions be represented as “x”, the formula becomes Ksp = x * x, simplifying to Ksp = x2.
Step 2: Calculating Molar Solubility
Provided that the Ksp of AgCl is given as 1.8 ‚àöo 10-10, we set up the equation x2 = 1.8 ‚àöo 10-10 to find “x”. Taking the square root gives us the molar solubility of Ag+, yielding a concentration of 1.3 ‚àöo 10-5 M. This concentration indicates how much AgCl can dissolve in pure water under equilibrium conditions.
Step 3: Impact of Common Ions on Solubility
When the initial concentration of Cl– ions is increased to 0.54 M (as in seawater), we must consider how this affects solubility. The presence of common ions leads to a common ion effect, which shifts the solubility equilibrium. By using an ICE table, the new molar solubility can be calculated, resulting in a decreased solubility of AgCl, noted as 3.3 ‚àöo 10-10 M for Cl ion concentration in seawater.