Question

1.Solid sodium sulfide is slowly added to 125 mL of a 0.0450 M silver nitrate solution. The concentration of sulfide ion required to just initiate precipitation is  M.

2.Solid barium acetate is slowly added to 50.0 mL of a 0.0522 M ammonium sulfite solution. The concentration of barium ion required to just initiate precipitation is  M.

3.Solid potassium hydroxide is slowly added to 150 mL of a 0.0329 M iron(III) nitrate solution. The concentration of hydroxide ion required to just initiate precipitation is  M.

Answer 1

-57 First we have to see in the reqeliom which salt is going to be precipitated We can get the required concentration of the ion by solving the equilibrio condition 9 = Ksp where Q = ionic product = product of concentration of ions of electrolyte , each raised to ksp= Aolubility product ibility product the power of their coefficients in balanced equation. į Sodium sulfide is added to 0.0450M Silver nitrate: 1. There will be precipitate of Silver Sulfide (Ag, 5) A 2Agit s² - ~ Aq, { (5) .: 9 = [ng?]*[se] [Ksp Dag 5 = 8 x 17" (standard value = 8x10 ftandard value from internet) e Tagalºus2ja 8x1054 at 25°C [52]= 8x10197 [494] = 0.095M TO:045]? = 3.95810-48 M The concentration of s2 should be 3.95810* lu for Ag, 5 precipitation Solid barium acetate is added to 0-05 22 ammonium sulfite - There will be precipitate of Barium sulfite (BaSO3) Ba?+ +5032 - Basoz 1:9= [Ba**] [50] [Kspleaso, 5x100 [ Bat"] [sos"]= 5 xloo [sog'] = 0.05 22M [Ba2+] = 5610 = 9.58X10-9 The concentration of Bat should be 9.58x10²M for Basoz precipitation -10 XID 0.05 22M 5 XIN JO 00522

2 032 pottasiumhydroxide is added to 0.0329M Fe(NO3), The precipate will be of Fe (OH)3 F22E+306° > Fe(OH)gt :: 9= [fe * ] [ou] [ "sp decom), * 4:87x16 17 -17 10 -17 [Febs] [ow]>= 487x10 11 [on'] = (427x600) TF2 393-0.0329 M 87x10-17 13 0.0329) = 1.19x105m - The concentration of one should be 114x10*m to jest preci pate Fc (04)3