The K s p of A g 2 C r O 4 , AgCl, AgBr and AgI are respectively, 1.1 × 1 0 - 12 , 1.8 × 1 0 - 10 , 5.0 × 1 0 - 13 , 8.3 × 1 0 - 17 . Which one of the following salts will precipitate last if A g N O 3 solution is added to the solution containing equal moles of NaCl, NaBr, NaI and N a 2 C r O 4 ?

Step-by-step solution

A g 2 C r O 4 1 1 - s → 2 A g + 0 2 s + C r O 4 - 2 0 s K s p = ( 2 s ) 2 s 1 - s s < < 1 As 1 >> s so 1 – s = 1 K s p = 4 s 3 = 1.1 × 1 0 - 12 = 4 s 3 = s = 6.5 × 1 0 - 5 A g C l → A g + s + C l - s 1.8 × 1 0 - 10 = s 2 or s = 1.34 × 1 0 - 5 Similarly 's' for AgBr and AgCl is found to be 7.1 × 1 0 - 7 and 9 × 1 0 - 9 respectively since solubility of A g 2 C r O 4 is highest, so it will participate last.