How much `NH_(3)`should be added to a solution of `0.01M Cu(NO_(3))_(2)` to reduce `[Cu^(2+)` to `10^(-13)`. Neglect the amount of copper in complexes containing fewer than `4` ammonia molecules per copper atom. Given `K_(d)` for `Cu(NH_(3))_(4)^(2+) = 1.0 xx 10^(-12)`

Explanation

To solve the problem of how much ammonia (NH₃) should be added to a solution of 0.01 M Cu(NO₃)₂ to reduce the concentration of Cu²⁺ ions to 10⁻¹³ M, we can follow these steps:

### Step 1: Write the dissociation equation for the complex
The dissociation of the complex ion Cu(NH₃)₄²⁺ can be represented as:
\[ \text{Cu(NH}_3\text{)}_4^{2+} \rightleftharpoons \text{Cu}^{2+} + 4 \text{NH}_3 \]

### Step 2: Write the expression for the dissociation constant (K_d)
The dissociation constant \( K_d \) for this reaction is given by:
...