Question

14-11 9. Using the RICE table, calculate the equilibrium concentration of each ion present when 0.5 g of PbCl is placed in water given the Ksp of PbCl2 is 1.60 x 10 at 20°C.

The question is asking to utilize the method of the RICE (or sometimes called just ICE) table

Answer 1

Podle is an insoluble ionic compound means it does not dissouate completely When place in aqueous solution. 80 an equilibrium is established in reaction which is gives kisp solubility Product mol of Pbcl2 = 0.5g - 0.00119 mol 278-18mort let the s amount is dissociated means solubility iss Pb (Op.) + 20 (ap) I (mol) Pollz () 0.00179 s S 25 25 Change At equilibrium is given as Ksp Pbul, = \-60x10-5 Solubility product Conslant Ksp Ksp = [Pb2+] [-] 1.6X10-5 = (s) (23) 1.6x10.5 = 453 0.0159 M - S Solubility J PbCl2 = 0.0159 At equilibrium [Pb2+] = 0.0159 m [u-) = 0.0318M

☺ bhg solution = 10 bola 14 - Du = 14-10 = 4 pok= -log [or] 4 = -log [ot] 10-4 = [ok] Concentration 9 Om = 1044 volume of solution = 546m2 ( mL = 0.0014 546 ml = 0.8462) molg or repuered = Molarity x volume () = 104mo) x 0.5546% = 5-46x10-5 mol BaCo4), → Barat) + 2017 Col.) 2 mol y 04- produced by 1 mol Bacon) 5:46X10-3 molg oh broduced by 2.73x10-5 mol Ba(04), | mol 1 Ba (H) = 273x/cº mol mang Ba (04)2 = mol x molar mass = 2.73x10-5 n900 x 169.346 g = 0.00469 mot man of Baloha = 0.00462 2 moljon