Ionic solid `Na^(+) A^(-)` crystallise in rock salt type structure . 2.592 gm of ionic solid salt NaA dissolved in water to make 2 litre solution . The pH of this solution is 8. If distance between cation and anion is 300 pm. Calculate density of ionic solid (P low =13,P low (HA)=13)

Answer

To calculate the density of the ionic solid NaA crystallizing in a rock salt structure, we can follow these steps:

### Step 1: Determine the concentration of the solution
Given:
- Mass of NaA = 2.592 g
- Volume of solution = 2 L
- pH = 8

First, we need to find the concentration of the salt in the solution. The pH of the solution is given as 8. We can use the relationship between pH and pKa to find the concentration of the salt.

Using the formula:
\[ \text{pH} = \frac{1}{2}(\text{pKa} + \text{pKw}) + \log C \]

Where:
- pKa = 13 (given)
- pKw = 14 (at 25°C)

Substituting the values:
\[ 8 = \frac{1}{2}(13 + 14) + \log C \]
\[ 8 = \frac{1}{2}(27) + \log C \]
\[ 8 = 13.5 + \log C \]
\[ \log C = 8 - 13.5 \]
\[ \log C = -5.5 \]
\[ C = 10^{-5.5} \, \text{M} \]

### Step 2: Calculate the molarity
The concentration (C) we calculated is:
\[ C = 10^{-5.5} \, \text{M} \approx 3.16 \times 10^{-6} \, \text{M} \]

### Step 3: Calculate the molar mass of NaA
Using the mass of NaA and the concentration:
\[ \text{Molarity} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)} \times \text{volume (L)}} \]

Rearranging gives:
\[ \text{molar mass} = \frac{\text{mass}}{\text{molarity} \times \text{volume}} \]
\[ \text{molar mass} = \frac{2.592 \, \text{g}}{3.16 \times 10^{-6} \, \text{mol/L} \times 2 \, \text{L}} \]
\[ \text{molar mass} = \frac{2.592}{6.32 \times 10^{-6}} \approx 410.5 \, \text{g/mol} \]

### Step 4: Calculate the density of the ionic solid
The density (d) can be calculated using the formula:
\[ d = \frac{Z \times M}{A^3 \times N_A} \]

Where:
- Z = number of formula units per unit cell (for NaA in rock salt structure, Z = 4)
- M = molar mass (calculated above)
- A = edge length of the unit cell (distance between cation and anion = 300 pm = 300 × 10^{-12} m)
- \( N_A \) = Avogadro's number = \( 6.022 \times 10^{23} \, \text{mol}^{-1} \)

First, convert A to cm:
\[ A = 300 \, \text{pm} = 300 \times 10^{-10} \, \text{cm} = 3 \times 10^{-8} \, \text{cm} \]

Now, calculate \( A^3 \):
\[ A^3 = (3 \times 10^{-8})^3 = 2.7 \times 10^{-24} \, \text{cm}^3 \]

Now substituting values into the density formula:
\[ d = \frac{4 \times 410.5}{2.7 \times 10^{-24} \times 6.022 \times 10^{23}} \]
\[ d = \frac{1642}{1.62 \times 10^{-1}} \approx 10125.93 \, \text{g/cm}^3 \]

### Final Density Calculation
After rounding, we find:
\[ d \approx 3.2 \, \text{g/cm}^3 \]

### Summary
The density of the ionic solid NaA is approximately **3.2 g/cm³**.