The molar solubility of silver bromide, AgBr in pure water is 0.0007350 mol/L. What is the Ksp for AgBr?
Express you answer in decimal form to 9 decimal places! Pls help

0.000000540

Explanation:

Step 1: Make an ICE chart for the solution of AgBr

"S" represents the molar solubility of AgBr

        AgBr(s) ⇄ Ag⁺(aq) + Br⁻(aq)

I                           0             0

C                          +S          +S

E                           S             S

Step 2: Write the expression for the solubility product constant (Ksp)

Ksp = [Ag⁺] [Br⁻] = S × S

Ksp = S² = (0.0007350)² = 0.000000540

5.6⋅10−^19  j

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