Answer to Question #333229 in Chemistry for Rick

Question

If The Ksp Of AgBr Is 5.0 X 10-13, What Is The Molar Solubility (Mol/L) Of AgBr?

Answer 1

Solution:

The K_sp of AgBr is 5.0×10⁻¹³

AgBr(s) ⇌ Ag⁺(aq) + Br⁻(aq)

The K_sp expression for AgBr(s) is:

K_sp = [Ag⁺][Br⁻]

ICE Table:

Substitute the equilibrium concentration values from the ICE Table into the K_sp expression: 

K_sp = [Ag⁺] × [Br⁻] = (x) × (x) = x² = 5.0×10⁻¹³

To find the molar solubility of AgBr, solve the equation for x:

x = (5.0×10⁻¹³)^1/2 = 7.07×10⁻⁷

Molar solubility of AgBr = 7.07×10⁻⁷ M

Answer: The molar solubility of AgBr is 7.07×10⁻⁷ mol/L