Question

A solution contains 0.021 M Cl? and 0.017 M I?. A solution containing copper (I) ions is added to selectively precipitate one of the ions. At what concentration of copper (I) ion will a precipitate begin to form? What is the identity of the precipitate? Ksp(CuCl) = 1.0 × 10-6, Ksp(CuI) = 5.1 × 10-12.

please show work!

A) 4 .8 × 10-5 M, CuCl
B) 3 .0 × 10-10 M, CuI
C) 3 .0 × 10-10 M, CuCl
D) 4 .8 × 10-5 M, CuI
E) N o precipitate will form at any concentration of copper (I).

Answer 1

The given K vales are as follows K for CuCl 1.0x106 K for CuI 5.1x102 The given concentrations of ions are as foll ows 0.017 м [cr-0021M Calculate the concentration of Cu+1 ion as follows +1 1.0x106 0.021 = 4.76x 10 M +1 5.1x10-2 0.017 - 3x10-10 M The concentrati on of of Cu ion required to form Culis less that of CuCl Therefore, from 3x10" M concentrati on of Cu*,the formation of precipitate starts to begin Therefore, the correct option is B) 3x10-10 M Cul