Determine the molar solubility of agbr in a solution containing 0.200 m nabr. ksp (agbr) = 7.7 * 10-13. determine the molar solubility of agbr in a solution containing 0.200 m nabr. ksp (agbr) = 7.7 * 10-13. 5.8 * 10-5 m 0.200 m 1.54 * 10-13 m 8.8 * 10-7 m 3.8 * 10-12 m

NaBr gives fully seperated ions as Na⁺ (aq) and Br⁻ (aq) while AgBr gives partially dissociated ions as Ag⁺ (aq) and Br⁻ (aq).

NaBr (aq) →Na⁺ (aq) + Br⁻ (aq)

0.200 0.200

AgBr (s) ⇄ Ag⁺ (aq) + Br⁻ (aq)

Initial - -

Change - X + X + X

Equilibrium X X

Ksp = [Ag⁺ (aq) ] [ Br⁻ (aq) ]

7.7 * 10⁻¹³ = X * (0.200 + X)

7.7 * 10⁻¹³ = X² + 0.200X

0 = X² + 0.200X-7.7 * 10⁻¹³

X1 = 3.8 x 10⁻¹²

X2 = - 0.2

X = molar solubility of AgBr.

hence, X cannot be a negative value.

molar solubility of AgBr is 3.8 x 10⁻¹² M.