Question

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

Answer 1

0.000000540

Step-by-step 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