Upon mixing equal volumes of 0 02 M AgNO3 0 16 M KCN

Last updated: 7/3/2023

Upon mixing equal volumes of 0 02 M AgNO3 0 16 M KCN solutions Assume no hydroly ion Given Ag aq 2CN aq Ag CN aq Kc 10 8 A NO 0 01 M K 0 08 M C Ag CN 0 01 M In the above question B 4 Ag is about A 2 78 x10 18 M C No free Ag ions will be left in the solution B CN 0 06 M D All of these B 1 67 x 10 1 M D Cannot be determined C