The formation constant* of [M(CN)6]^4– is 2.50 × 10^17, where M is a generic metal. A 0.130-mole quantity of M(NO3)2 is added to a liter of 1.170 M NaCN solution. What is the concentration of M^2+ ions at equilibrium?
Answer:
M^2+(aq) + 6CN^- ⇋ [M(CN)6]^4-
Kform = [[M(CN)6]^4-]/[ M^2+][CN^-]^6 = 2.50×10^17
After equilibrium established let [M^2+] = x M
[[M(CN)6]^4-] = (0.170)- x ≈ 0.170 as we shall see x is minute and
this approximation is justified.
Uncomplexed [CN^-] = 1.28 - 6×0.170 = 0.26 M
Put these values into Kform:
0.170/x(0.26)^6 = 2.50×10^17
x = 0.170/[(2.50×10^17)×(0.26)^6] = 2.20×10^-15 M
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