Question

The Kp for lead iodide (Pbl) is 1.4 x 108. Calculate the solubility of lead iodide in each of the following a. water Solubility = mol/L b. 0.22 M Pb(NO3)2 Solubility mol/L c. 0.022 M Nal Solubility = mol/L Submit Answer Try Another Version 6 item attempts remaining

Answer 1

pbI & pbt² + 21 s 25 sa molar solubility nt Ksp = [ Pb+e] L1"] Espaz (3) (23)+= 45? . esp = 4.83 nt کے 1-4X152 = 45 HUXLOP 20.35 Y3 0-3333 2 (0-35 X10°) so (0.35 X10%) 1.51928x10 Szo.00151928 molla

6 Pb (NO3)₂ → Pyth + 2NO [P672] = 0.22M pb Į t pbt2 + 2I' 22 +x 0.22 +x +2x 24 : Isp = [P6+4] [15] Esp = (0.22 +x) ( 2x) For sparingly soluble salt's 0-22+x 0.22 so Esp = 0.22 x 44² 1.4 Xlorea 0.88x² 2 1.4 X10 h X 0.88 1.48108 0.88 . molar solubility 20.000 126 1312 xz 1.261312 x 104 X= 1-26 13k X 104 mo! L

NaI → Nat & I [I-) = 0.022M pbą o p6th + 2I - To 10.022 -X +* +-X = x 70.022+28 X Ksp = [pft"] [I.]? Ksp 2 x Co.oze + 2x) For sparingly soluble salt's 0.022+ 2x B 0-022 so Esp = x (0.022) 1.4x108 X (0.02237 14x108 X= 10.022) Xa o. 0000 289256 x= 2189256 xlos mol