Question

Given that, for Agi, at 25°C, Ksp = 1.5x10-16: What is the solubility (g/mL) of Agi in water at 25°C? Submit Answer Tries 0/5 20.0 mL of 0.060 M Nal is mixed with 20.0 mL of 0.120 M AgNO3. What is [I"] in the supernatant solution at equilibrium, at 25°C? Submit Answer Tries 0/5 What is the solubility (mass/volume) of AgI in this supernatant solution? Submit Answer Tries 0/5

Answer 1

Silver Iodide (Ag I) will dissociate water into its constituent ions. Agt and I The equation representing this dissociation is AgI a Ag+ + I So I mole of Ag I will into Imole of Ag and DO dissociate Imole of Let 's' be the solubility of each of Ag and I ions. Since initially we only have solid Ag I and no Ag+ and I- ions so initial concentrato of ions is zero. Thus ICE table this equilibrium Ag I Ag+ + I Initial (M) Change (M) Equilibrium (M) + b Since solubility expressed as the concentrations of product (Ksp) is product of equilibrium product ions so we have

Ksp = [Ag+][I ] =0 where [Ag+] and [I] represent equilibrium concentrations of Ag. -Iions Giren Ksp for Ag I = 1.58 10-16. Also from Ice table at equilchen [Ag+] = [1-325 So equation becomes . 1.5. X10-16 = Sxs 15x10-16 This .. s= S r.sx 10-16 1.22 x 10-8 M This concentration at equilibrium is of Ag+ and I 1922 x 10-8 M... ions Molar concentration or molar solubility calculated above can be converted to solubility by using molar mass of

Solubility = 1:22x 10-mol x 234.774 L - mol (Because M = mol/L is the unit of -molarityanand molar mass of is equal to 234.77 g/mol) bubility = 1?2x10- yael X.234.779 - L mot = 1:22X10-8 x 234.774 =-1.22.X.234.77X10-84 1000 mL (Because I L=1000m = 1.22%234.71x10- amb 103 = 286.42 x 10-8x10-3 gimi = 286.42 x 10-8+(-3) 3 286.42X10" almi - 12.86 x 10-9 ~ 2.9 x 10-9 g/mL

Thus 2.9 × 10^-9 g of AgI can dissolve in 1 mL of water , which means the solubility of AgI in water at 25°C is 2.9 × 10^-9 g/mL.