`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)`
Edge length of FCC unit cell of copper sulphate is

Options

A. 120 Å
B. 170 Å
C. 150 Å
D. 200 Å

Explanation

To solve the problem, we need to determine the edge length of the FCC unit cell of copper sulfate (CuSO₄) based on the information provided. Here’s a step-by-step breakdown of the solution:

### Step 1: Determine the concentration of Cu²⁺ ions
The equilibrium reaction provided is:
\[ \text{Cu(H}_2\text{O})_6^{2+} + \text{H}_2\text{O} \rightleftharpoons \text{[Cu(H}_2\text{O})_5\text{(OH)}]^+ + \text{H}_3\text{O}^+ \]
Given that the equilibrium constant \( K = 10^{-5} \), we can set up the expression for the equilibrium concentration.

Let the initial concentration of Cu²⁺ ions be \( C \) and the change in concentration at equilibrium be \( x \). Thus, at equilibrium:
- Concentration of \(\text{Cu(H}_2\text{O})_6^{2+}\) = \( C - x \)
- Concentration of \(\text{[Cu(H}_2\text{O})_5\text{(OH)}]^+\) = \( x \)
- Concentration of \(\text{H}_3\text{O}^+\) = \( x \)

From the equilibrium expression:
\[ K = \frac{x^2}{C - x} = 10^{-5} \]

### Step 2: Substitute the pH to find \( x \)
Given that the pH of the solution is 5, we can find the concentration of \(\text{H}_3\text{O}^+\):
\[ \text{pH} = -\log[\text{H}_3\text{O}^+] \]
\[ [\text{H}_3\text{O}^+] = 10^{-5} \, \text{mol/L} \]
Thus, \( x = 10^{-5} \).

### Step 3: Substitute \( x \) into the equilibrium expression
Substituting \( x \) into the equilibrium expression:
\[ 10^{-5} = \frac{(10^{-5})^2}{C - 10^{-5}} \]
This simplifies to:
\[ C - 10^{-5} = 10^{-5} \]
Thus,
\[ C = 2 \times 10^{-5} \, \text{mol/L} \]

### Step 4: Calculate the moles of CuSO₄ dissolved
The volume of the solution is 500 mL, which is 0.5 L. Therefore, the moles of CuSO₄ dissolved is:
\[ \text{Moles of CuSO}_4 = C \times \text{Volume} = 2 \times 10^{-5} \, \text{mol/L} \times 0.5 \, \text{L} = 10^{-5} \, \text{mol} \]

### Step 5: Calculate the number of unit cells
In a face-centered cubic (FCC) structure, there are 4 formula units per unit cell. The number of unit cells is given by:
\[ \text{Number of unit cells} = \frac{\text{Moles of CuSO}_4 \times N_A}{4} \]
Where \( N_A \) (Avogadro's number) is approximately \( 6.022 \times 10^{23} \):
\[ \text{Number of unit cells} = \frac{10^{-5} \times 6.022 \times 10^{23}}{4} = 1.5 \times 10^{19} \]

### Step 6: Calculate the number of unit cells along one edge of the cube
The number of unit cells along one edge of the cube is:
\[ n = \sqrt[3]{1.5 \times 10^{19}} \approx 1.14 \times 10^6 \]

### Step 7: Calculate the edge length of the unit cell
The total edge length of the cubic crystal is given as 17.1 mm (or 17100 µm). The edge length \( a \) of the unit cell can be calculated as:
\[ a \times n = 17100 \, \mu m \]
\[ a = \frac{17100 \, \mu m}{1.14 \times 10^6} \approx 0.015 \, \mu m = 150 \, \text{Å} \]

### Final Answer
The edge length of the FCC unit cell of copper sulfate is approximately **150 Å**.

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