Chemical EquilibriumHard
Question
The formation constant of Ni(NH3)62+ is 6 × 108 at 250C. If 50 ml of 2.0 M NH3 is added to 50 ml of 0.20 M solution of Ni2+, the concentration of Ni2+ ion will be nearly equal to :
Options
A.3 × 10-10 mole litre-1
B.2 × 10-10 mole litre-1
C.2 × 10-9 mole litre-1
D.4 × 10-8 mole litre-1
Solution
Ni+2 + 6NH3 → [Ni (NH3)]+6 Kf = 6 × 108
t = 0 0.01 mole 0.1 mole 0
Kc =
= 6 × 108. [Ni+2] = 4 × 10-8.
t = 0 0.01 mole 0.1 mole 0
Kc =
= 6 × 108. [Ni+2] = 4 × 10-8.Create a free account to view solution
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