Can someone help me answer this please

A) 1.3 × 10⁻⁵ mol/L

Explanation:

Step 1: Write the balanced equation for the solution of AgCl

AgCl(s) ⇄ Ag⁺(aq) + Cl⁻(aq)

Step 2: Make an ICE Chart

        AgCl(s) ⇄ Ag⁺(aq) + Cl⁻(aq)

I                           0             0

C                        +S            +S

E                          S              S

If we replace the solubility (S) in the Ksp expression, we get,

Ksp = [Ag⁺] [Cl⁻] = S × S = S²

S = √Ksp = √1.8 × 10⁻¹⁰ = 1.3 × 10⁻⁵ mol/L

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