What is `[Cd^(2+)]` in `1.0L` of solution prepared by dissolving `0.001mol Cd(NO_(3))_(2)` and `1.5 mmol. NH_(3) ? K_(d)` for the dissociation of `Cd(NH_(3))_(4)^(2+)` into `Cd^(2+)` and `4NH_(3)` is `1.8 xx 10^(-7)`. Neglect the amount of `Cd` in complexes containing fewer than `4` ammonia molecules.

Explanation

To solve the problem of finding the concentration of `[Cd^(2+)]` in a solution prepared by dissolving `0.001 mol Cd(NO3)2` and `1.5 mmol NH3`, we will follow these steps:

### Step 1: Write the equilibrium expression for the dissociation of the complex
The dissociation of the complex ion `Cd(NH3)4^(2+)` can be represented as:
\[ Cd(NH3)_4^{2+} \rightleftharpoons Cd^{2+} + 4NH3 \]

### Step 2: Set up the initial concentrations
Initially, we have:
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