`CuSO_4` crystallise in rock salt structure.Its cell parameter can be determined by various experimental methods like electrical conductivity measurement, colligative properties measurement, pH measurement etc.
A cubic crystal of `CuSO_4` of edge length 17.1 mm, is dissolved in water to make 500 ml solution of pH 5.`root3(1.5)=1.14)`
Given : `Cu(H_2O)_6^(2+)+H_2O hArr [Cu(H_2O)_5(OH)]^(+)+H_3O^(+) , K=10^(-5)`
In the given solution is made 1 M with respect to `[Cu^(2+)]` becomes `10^(-15)` M therefore `K_f` for the formation of `Cu(NH_3)_4^(2+)` is

Explanation

To solve the problem, we need to determine the formation constant \( K_f \) for the complex ion \( \text{Cu(NH}_3\text{)}_4^{2+} \) based on the given information. Here’s a step-by-step breakdown of the solution:

### Step 1: Determine the concentration of \( \text{Cu}^{2+} \)
The problem states that the pH of the solution is 5. Therefore, the concentration of \( \text{H}_3\text{O}^+ \) ions can be calculated as follows:

\[ [\text{H}_3\text{O}^+] = 10^{-pH} = 10^{-5} \text{ M} \]

### Step 2: Use the equilibrium expression for the reaction
The equilibrium reaction given is:

\[ \text{Cu(H}_2\text{O)}_6^{2+} + \text{H}_2\text{O} \rightleftharpoons \text{Cu(H}_2\text{O)}_5\text{OH}^{+} + \text{H}_3\text{O}^{+} \]

The equilibrium constant \( K \) for this reaction is given as \( 10^{-5} \).

### Step 3: Set up the equilibrium expression
Let the initial concentration of \( \text{Cu(H}_2\text{O)}_6^{2+} \) be \( C \) and the change in concentration at equilibrium be \( x \). Thus, we can express the equilibrium concentrations as:

- \( [\text{Cu(H}_2\text{O)}_6^{2+}] = C - x \)
- \( [\text{Cu(H}_2\text{O)}_5\text{OH}^{+}] = x \)
- \( [\text{H}_3\text{O}^{+}] = 10^{-5} \)

Substituting into the equilibrium expression:

\[ K = \frac{x \cdot 10^{-5}}{C - x} = 10^{-5} \]

### Step 4: Solve for \( x \)
Assuming \( x \) is small compared to \( C \), we can approximate \( C - x \approx C \):

\[ 10^{-5} = \frac{x \cdot 10^{-5}}{C} \]

This simplifies to:

\[ x = C \]

### Step 5: Calculate the concentration of \( \text{Cu}^{2+} \)
From the problem, we know that when the solution is made 1 M with respect to \( [\text{Cu}^{2+}] \), it becomes \( 10^{-15} \) M. Therefore:

\[ C = 10^{-15} \text{ M} \]

### Step 6: Calculate the formation constant \( K_f \)
The formation constant \( K_f \) for the complex \( \text{Cu(NH}_3\text{)}_4^{2+} \) can be expressed as:

\[ K_f = \frac{[\text{Cu(NH}_3\text{)}_4^{2+}]}{[\text{Cu}^{2+}][\text{NH}_3]^4} \]

Given that \( [\text{Cu}^{2+}] = 10^{-15} \) M and \( [\text{Cu(NH}_3\text{)}_4^{2+}] = 2 \times 10^{-5} \) M (from the previous calculations):

\[ K_f = \frac{2 \times 10^{-5}}{(10^{-15}) \cdot (1)^4} \]

### Step 7: Final calculation
Calculating \( K_f \):

\[ K_f = \frac{2 \times 10^{-5}}{10^{-15}} = 2 \times 10^{10} \]

### Conclusion
Thus, the formation constant \( K_f \) for the formation of \( \text{Cu(NH}_3\text{)}_4^{2+} \) is:

\[ \boxed{2 \times 10^{10}} \]