[M2+] added = number of moles / volume
                  Â
= 0.13 / 1
                  Â
= 0.13 M
[CN-] added = 1.15 M
M2+Â Â Â Â Â
+Â Â Â Â Â Â Â
6CN-Â Â Â Â Â Â Â --------->
[M(CN)6]4–
0.13Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â
1.15Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â
0Â Â Â Â
Since Kf is very large, reaction will go to
completion
From reaction, 1 mol of M2+ reacts with 6 mol og CN-
so, 0.13 mol of M2+ requires= 0.13*6 = 0.78 mol of
CN-
So M2+ is limiting reagent and whole of M2+ will
react
So,
there will not be any M2+ at equilibrium
