Ionic solid `B^+A^(-)` crystallizes in rock salt type of structure.1.296 gm ionic solid salt `B^+A^(-)` is dissolved in water to make one litre solution.The pH of the solution is measured to be 6.0.if the value of face diagonal in the unit cell of `B^+A^(-)` be `600sqrt2` pm.Calculate the density of ionic solid in gm/cc.`(T=298 K, K_b "for" BOH "is" 10^(-6)`,(Avogadro Number =`6.0xx10^(23)`))

Explanation

To solve the problem step by step, we will follow the information provided in the question and the video transcript.

### Step 1: Calculate the concentration (C) of the solution
Given:
- pH = 6.0
- pKw = 14 (at 298 K)

Using the relation:
\[ \text{pH} + \text{pOH} = \text{pKw} \]
We can find pOH:
\[ \text{pOH} = 14 - \text{pH} = 14 - 6 = 8 \]

Now, we can find the concentration (C) using the formula:
\[ \text{pOH} = \frac{1}{2} (\text{pKw} - \text{pKb} - \log C) \]
Rearranging gives:
\[ \log C = \text{pKw} - \text{pKb} - 2 \cdot \text{pOH} \]

Given \( K_b \) for \( BOH = 10^{-6} \):
\[ \text{pKb} = -\log(10^{-6}) = 6 \]

Now substituting the values:
\[ \log C = 14 - 6 - 2 \cdot 8 \]
\[ \log C = 14 - 6 - 16 = -8 \]
Thus,
\[ C = 10^{-8} \, \text{mol/L} \]

### Step 2: Calculate the molar mass of the ionic solid
We know:
\[ C = \frac{1.296 \, \text{g}}{M} \]
Where M is the molar mass. Rearranging gives:
\[ M = \frac{1.296 \, \text{g}}{C} = \frac{1.296 \, \text{g}}{10^{-8} \, \text{mol/L}} = 1.296 \times 10^{8} \, \text{g/mol} \]

### Step 3: Calculate the edge length (a) of the unit cell
Given the face diagonal:
\[ d = 600\sqrt{2} \, \text{pm} = 600\sqrt{2} \times 10^{-10} \, \text{cm} \]
For a cubic unit cell, the relationship between the face diagonal and the edge length is:
\[ d = a\sqrt{2} \]
Thus,
\[ a = \frac{d}{\sqrt{2}} = \frac{600\sqrt{2} \times 10^{-10}}{\sqrt{2}} = 600 \times 10^{-10} \, \text{cm} = 6 \times 10^{-8} \, \text{cm} \]

### Step 4: Calculate the density of the ionic solid
For rock salt structure, the number of formula units (Z) is 4. The density (ρ) can be calculated using the formula:
\[ \rho = \frac{Z \cdot M}{N_A \cdot a^3} \]
Where:
- \( N_A = 6.022 \times 10^{23} \, \text{mol}^{-1} \)
- \( a^3 = (6 \times 10^{-8} \, \text{cm})^3 = 2.16 \times 10^{-23} \, \text{cm}^3 \)

Substituting the values:
\[ \rho = \frac{4 \cdot (1.296 \times 10^{8})}{6.022 \times 10^{23} \cdot 2.16 \times 10^{-23}} \]
Calculating this gives:
\[ \rho \approx 4 \, \text{g/cm}^3 \]

### Final Answer
The density of the ionic solid \( B^+A^- \) is approximately **4 g/cm³**.

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