Question

Lunes CHEMWORK The Hg2+ ion forms complex ions with I as follows: Hg2+ + HgT+ HgI + 1 =Hg2 K1 = 1.0 108 K2=1.0 x 105 K3 -1.0 x 10° K4 = 1.0 * 108 Hgl2 + 1 + HgI3 HgI3 + 1 = Hg142 A solution is prepared by dissolving 0.057 mol Hg(NO3)2 and 5.00 mol Nal in enough water to make 1.0 L of solution. Calculate the equilibrium concentration of [Hg142]. (Hg142) = M Calculate the equilibrium concentration of [I'). Calculate the equilibrium concentration of [Hg2+). [Hg2+1 =

Answer 1

Comment in case of any doubt.

180= k, = [Hadit] since equilibrium constant of each of the reactions is very high, this each reaction occurs to a very high extent. - Combining all the given equations, we get the net equation as follows: ICE Table Hight + ur = Hale keq = kxket Kgx ky I: 0.057 9.00 - jo sjók vox) of c: -M -41 +01 = 180 E: (0.059-») (5-4) on ! Key were in any = 6.057)S-vou) () 1. Since Keg is very high so linuting reactant will be almost fully consumed (as reaction will almost go to completion ). - For reactant Hight, moles of Nal required to react fully with Hg2+ =.0057+ 4 = 0.22 smal 1. This, an 0.228mal ... [17] = 5-44= 5 -0.228 = 4.772 mal, using as above, we get 100-10.057 _= Chetting 1 [Hg2+] 1 44.772) = 1.1981073 Thus, Inga] = 1.1 axit h 10.057M [1] = [4177] M [Higar) = 119x 1032m] 704)15-401)