Calculate the pH of solution prepared by dissolving 2 g of TlOH in 2 L of solution.

Explanation

To calculate the pH of a solution prepared by dissolving 2 g of TlOH in 2 L of solution, follow these steps:

### Step 1: Calculate the number of moles of TlOH
First, we need to find the molar mass of TlOH (Thallium hydroxide). The molar mass can be calculated as follows:
- Molar mass of Thallium (Tl) = 204.38 g/mol
- Molar mass of Oxygen (O) = 16.00 g/mol
- Molar mass of Hydrogen (H) = 1.01 g/mol

So, the molar mass of TlOH = 204.38 + 16.00 + 1.01 = 221.39 g/mol.

Now, calculate the number of moles of TlOH:
\[ \text{Number of moles} = \frac{\text{Given weight}}{\text{Molar mass}} = \frac{2 \, \text{g}}{221.39 \, \text{g/mol}} \approx 0.00904 \, \text{mol} \]

### Step 2: Calculate the concentration of TlOH in the solution
Concentration (C) is calculated using the formula:
\[ C = \frac{\text{Number of moles}}{\text{Volume in liters}} \]
Given that the volume of the solution is 2 L:
\[ C = \frac{0.00904 \, \text{mol}}{2 \, \text{L}} \approx 0.00452 \, \text{mol/L} \]

### Step 3: Determine the concentration of hydroxide ions (OH⁻)
Since TlOH is a strong base, it dissociates completely in solution:
\[ \text{TlOH} \rightarrow \text{Tl}^+ + \text{OH}^- \]
This means that the concentration of OH⁻ ions will also be 0.00452 mol/L.

### Step 4: Calculate pOH
The pOH can be calculated using the formula:
\[ \text{pOH} = -\log[\text{OH}^-] \]
Substituting the concentration of OH⁻:
\[ \text{pOH} = -\log(0.00452) \approx 2.35 \]

### Step 5: Calculate pH
Using the relationship between pH and pOH:
\[ \text{pH} + \text{pOH} = 14 \]
We can find the pH:
\[ \text{pH} = 14 - \text{pOH} = 14 - 2.35 \approx 11.65 \]

### Final Answer
The pH of the solution is approximately **11.65**.