5.35 g of a salt Acl (of weak base AOH) is dissolved in 250 ml of solution. The pH of the resultant solution was found to be 4.827. Find the ionic radius of `A^(+)` and `Cl^(-)` if ACl forms CsCl type crystal having 2.2 g/cm3. Given `K_(b(AOH))=1.8xx10^(-5), (r^(+))/(r^(-))=0.732` for this cell unit.

Answer

To solve the problem step by step, we will follow the outlined process:

### Step 1: Calculate pKb
Given the base dissociation constant \( K_b \) for AOH:
\[ K_b = 1.8 \times 10^{-5} \]
To find \( pK_b \):
\[ pK_b = -\log(K_b) = -\log(1.8 \times 10^{-5}) \approx 4.74 \]

### Step 2: Calculate pH and pKa
Given the pH of the solution:
\[ \text{pH} = 4.827 \]
Using the relationship:
\[ pK_w = 14 \quad \text{and} \quad pK_a + pK_b = pK_w \]
We find \( pK_a \):
\[ pK_a = pK_w - pK_b = 14 - 4.74 \approx 9.26 \]

### Step 3: Calculate the concentration (C) of the salt ACl
Using the formula for pH:
\[ \text{pH} = \frac{1}{2}(pK_a + pK_w - pK_b - \log C) \]
Rearranging gives:
\[ \log C = pK_a + pK_w - pK_b - 2 \times \text{pH} \]
Substituting the known values:
\[ \log C = 9.26 + 14 - 4.74 - 2 \times 4.827 \]
Calculating:
\[ \log C = 9.26 + 14 - 4.74 - 9.654 \approx 8.866 \]
Thus,
\[ C \approx 0.4 \, \text{mol/L} \]

### Step 4: Calculate the molar mass of ACl
Using the formula for molarity:
\[ \text{Molarity} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)} \times \text{volume (L)}} \]
Given mass = 5.35 g and volume = 0.250 L:
\[ 0.4 = \frac{5.35}{\text{molar mass} \times 0.250} \]
Rearranging gives:
\[ \text{molar mass} = \frac{5.35}{0.4 \times 0.250} = 53.5 \, \text{g/mol} \]

### Step 5: Calculate the edge length (a) of the unit cell
Using the density formula:
\[ \text{Density} = \frac{n \cdot M}{N_A \cdot a^3} \]
Where:
- \( n = 1 \) (for CsCl type structure)
- \( M = 53.5 \, \text{g/mol} \)
- \( N_A = 6.022 \times 10^{23} \, \text{mol}^{-1} \)
- Density = 2.2 g/cm³

Rearranging gives:
\[ a^3 = \frac{n \cdot M}{\text{Density} \cdot N_A} \]
Substituting values:
\[ a^3 = \frac{1 \cdot 53.5}{2.2 \cdot 6.022 \times 10^{23}} \approx 3.43 \times 10^{-22} \, \text{cm}^3 \]
Calculating \( a \):
\[ a \approx 3.43 \times 10^{-8} \, \text{cm} \]

### Step 6: Calculate the ionic radii
Using the relationship:
\[ r^+ + r^- = \frac{\sqrt{3}}{2} a \]
Substituting for \( a \):
\[ r^+ + r^- = \frac{\sqrt{3}}{2} \cdot 3.43 \times 10^{-8} \approx 2.97 \times 10^{-8} \, \text{cm} \]

Given the ratio \( \frac{r^+}{r^-} = 0.732 \):
Let \( r^- = r \):
\[ r^+ = 0.732r \]
Substituting into the equation:
\[ 0.732r + r = 2.97 \times 10^{-8} \]
\[ 1.732r = 2.97 \times 10^{-8} \]
Calculating \( r \):
\[ r \approx 1.71 \times 10^{-8} \, \text{cm} \]
Then:
\[ r^+ \approx 1.26 \times 10^{-8} \, \text{cm} \]

### Final Results
- Ionic radius of \( A^+ \) (cation) = \( 1.26 \times 10^{-8} \, \text{cm} \)
- Ionic radius of \( Cl^- \) (anion) = \( 1.71 \times 10^{-8} \, \text{cm} \)