Question

A solution contains 1.48x102 M zinc nitrate and 9.13*10' M lead acetate. Solid potassium sulfide is added slowly to this mixture. What is the concentration of lead ion when zinc ion begins to precipitate? [Pb2') = M

Answer 1

of Given information : - Concentration rine nitrate = 1.48x 10²m concentration of lead acetate = 9.13x103m. If potassium sulfide is added to the mixture then precipitates of zine sulfide and lead sulfides formed . are -25 ksip of Pbs first . to Ksp of ins = 2x10 9.01x10-29 kesp of pbs is lower than ans. So, abs & precipitates The calculation of sulfide ion which is required precipitate zet is shown below:- [Zn2+)? 14.8x16M [52] = [znet] kop 2x10-25 2 1.48x10-2 1.35x10 -23 M

27 of pb 9. 01x 10.29 The calculation of concentration s shown belowi- [P12+] = kap [s2] 1.35x10 = 6.67x10-6M Therefore, the concentration Pb2+ ion when anation begin to Precipitate is 6.6781006M -23 of