Tritium, T(an isotope of H) combines with fluorine to form weak acid, TF, which ionizes to give T'. Tritium is radioactive and is a `beta`-emitter. A freshly prepared aqueous solution of TF has pT (equialent of pH) of `1.5` and freezes at `-0.372^(@)C`. If 600mL of freeshly prepared solution were allowed to stad for `24.8` years, calculate (i) ionization constant of TF (ii) Number of `beta-`particles emitted.
(Given `K_(f)` for water `= 1.86 kg mol K^(-1) t_(1//2)` for tritium `= 12.4` years).

Explanation

To solve the problem, we will break it down into two parts: (i) calculating the ionization constant (Ka) of the weak acid TF, and (ii) calculating the number of beta particles emitted after 24.8 years.

### Part (i): Calculation of Ionization Constant (Ka)

1. **Determine the concentration of T⁺ ions:**
Given that the pT (equivalent to pH) is 1.5, we can find the concentration of T⁺ ions:
\[ [T^+] = 10^{-pT} = 10^{-1.5} = 0.0316 \, \text{mol/L} \]
Let this be equation (1).

2. **Define the dissociation of TF:**
The weak acid TF dissociates as follows:
\[ TF \rightleftharpoons T^+ + F^- \]
If the initial concentration of TF is C and the degree of ionization is α, then:
- Concentration of TF at equilibrium = \( C(1 - \alpha) \)
- Concentration of T⁺ = \( C\alpha \)

From the dissociation, we know:
\[ [T^+] = C\alpha \]
Thus, from equation (1):
\[ C\alpha = 0.0316 \, \text{mol/L} \quad \text{(equation 2)} \]

3. **Calculate freezing point depression (ΔTf):**
The freezing point depression is given as -0.372 °C. The formula for freezing point depression is:
\[ \Delta T_f = I \cdot K_f \cdot m \]
Where:
- \( K_f = 1.86 \, \text{kg mol}^{-1} \)
- \( m \) is the molality, which is approximately equal to the molarity for dilute solutions.

Since TF dissociates into two ions (T⁺ and F⁻), the van 't Hoff factor (I) is:
\[ I = 1 + (n - 1)\alpha = 1 + \alpha \quad (n = 2) \]
Therefore, we can write:
\[ 0.372 = (1 + \alpha) \cdot 1.86 \cdot C \]
Let this be equation (3).

4. **Solve equations (2) and (3):**
From equation (2):
\[ C = \frac{0.0316}{\alpha} \]
Substitute this into equation (3):
\[ 0.372 = (1 + \alpha) \cdot 1.86 \cdot \left(\frac{0.0316}{\alpha}\right) \]
Rearranging gives:
\[ 0.372\alpha = (1 + \alpha) \cdot 1.86 \cdot 0.0316 \]
Solving this equation will yield values for C and α.

5. **Calculate Ka:**
The ionization constant (Ka) is given by:
\[ K_a = \frac{[T^+][F^-]}{[TF]} = \frac{C\alpha \cdot C\alpha}{C(1 - \alpha)} = \frac{C\alpha^2}{1 - \alpha} \]
Substitute the values of C and α obtained from the previous steps to find Ka.

### Part (ii): Calculation of Number of Beta Particles Emitted

1. **Calculate initial moles of TF:**
The volume of the solution is 600 mL, which is 0.6 L. Using the concentration C from previous calculations:
\[ \text{Initial moles of TF} = C \times 0.6 \]

2. **Determine remaining moles after 24.8 years:**
The half-life of tritium (t₁/₂) is 12.4 years. After 24.8 years, which is two half-lives:
\[ \text{Remaining moles} = \frac{\text{Initial moles}}{2^2} = \frac{\text{Initial moles}}{4} \]

3. **Calculate moles disintegrated:**
\[ \text{Moles disintegrated} = \text{Initial moles} - \text{Remaining moles} \]

4. **Calculate the number of beta particles emitted:**
Each disintegration emits one beta particle. Therefore, the number of beta particles emitted is:
\[ \text{Number of beta particles} = \text{Moles disintegrated} \times N_A \]
Where \( N_A \) is Avogadro's number (\( 6.022 \times 10^{23} \, \text{particles/mol} \)).