Answer
$1 Cd^{2+}(aq) + 4 NH_3(aq) \lt -- \gt 1 [Cd(NH_3)_4]^{2+}$
$K_f = \frac{[[Cd(NH_3)_4]^{2+}]^1}{[Cd^{2+}]^1[NH_3]^4}$
Work Step by Step
1. Identify the separate compounds and their quantities in this complex ion:
- 1 $Cd^{2+}$
- 4 $NH_3$
2. Write the equation where they combine to formate the complex ion $[Cd(NH_3)_4]^{2+}$
$1 Cd^{2+}(aq) + 4 NH_3(aq) \lt -- \gt 1 [Cd(NH_3)_4]^{2+}$
3. Write the formation constant expression:
- It will be: the concentrations of the products divided by that of the reactants:
- And remember: the balance coefficients will be the exponents of their concentrations.
$K_f = \frac{[Products]}{[Reactants]}$
$K_f = \frac{[[Cd(NH_3)_4]^{2+}]^1}{[Cd^{2+}]^1[NH_3]^4}$
