solubility product of AgCl is 1.77 * (10^ -9) in 25 degrees C. Calculate the solubility of Ag+ ion in 0.1M NaCl solution!?

0.1 M NaCl completely dissociates in aqueous solution to give 0.1 M HCl.

____________ AgCl(aq) __ ⇌ __ Ag⁺(aq) __ + __ Cl⁻(aq) ___ Ksp = 1.77 × 10⁻⁹
Initial: _____________________ 0 M _________ 0.1 M
Change: ___________________ +s M _________ +s M
At eqm: ____________________ s M _______ (0.1 + s) M

As Ksp is very small and common ion effect due to the presence of Cl⁻ ions , the solubility is very small.
It is assumed that 0.1 ≫ s, i.e. [Cl⁻] at equilibrium = (0.1 + s) M ≈ 0.1 M

Ksp = [Ag⁺] [Cl⁻]
1.77 × 10⁻⁹ = s × 0.1
s = (1.77 × 10⁻⁹) / 0.1
s = 1.77 × 10⁻⁸ (This agrees to the assumption that 0.1 ≫ s.)

Solubility of Ag⁺ ion = 1.77 × 10⁻⁸ M