Solution :
The equilibrium reaction and ICE table can be given as,
M^2+ + 2SCN- <--> M(SCN)2
0.000100 M --- 0.000100 M ---- 0 M (initial)
-X ---------------- -2X -------------- X (change)
(0.000100 - X) --- (0.000100 -X) --- X (equilibrium)
Given, equilibrium concentration of M(SCN)2 = X = 1.0 x 10^-5 M
Hence,
Equilibrium concentration of M2+ ion = 0.000100 - 1.0 x 10^-5
= 10 x 10^-5 - 1.0 x 10^-5 = 9.0 x 10^-5 M
Equilibrium concentration of SCN- ion = 0.000100 - 2 x 1.0 x 10^-5
= 10 x 10^-5 - 2 x 10^-5 = 8.0 x 10^-5 M
Thus,
[M^2+] = 9.0 x 10^-5 M
[SCN-] = 8.0 x 10^-5 M
Kc = [M(SCN)2] / [M2+] [SCN-]^2
Kc = 1.0 x 10^-5 / 9.0 x 10^-5 x (8.0 x 10^-5)^2
Kc = 1.74 x 10^7
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