Eq constant for reaction
[Ag(CN)2]- ≈ Ag+ +2CN- is 4 x 10-19
calulate silver ion conc in a soln which is originally 0.1M in KCN and 0.03 M in AgNO3
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1 Answers
12CN- + Ag+ → [Ag(CN)2]+
at t=0 0.1M 0.03M 0
at equi. 0.1-0.06 0 0.03M
= 0.04M
I n this react, conc. of Ag+ is zero
so equil. is shifted towards left.
hence by dissociation some Ag+ is formed again
[Ag(CN)2]+ → Ag+ + 2CN-
at equilibruim 0.03 0 0.04
at new equlibrium 0.03 -x x 0.04+2x
K = [Ag+][CN- ]2[Ag(CN)2]+
hence 4 x 10-19 = x (0.04+2x)0.03-x
we can take 0.03 - x ≈ 0.03 since x wuld be small.
we get by neglecting further higher power of x as x = 7.5 x 10-18M
hence [Ag+] = 7.5 x 10-18M
