Equilibrium constants of `T_(2)O` `(T or _(1)^(3)H` is an isotope `of _(1)^(1)H)`and `H_(2)O` are different at 298 K. Let at 298 K pure `T_(2)O` has pT (like pH) is 7.62. The pT of a solution prepared by adding 10 mL. of 0.2 M TCl to 15 mL of 0.25 M NaOT is:

Options

A. `2-log 7`
B. `14+log 7`
C. `13.24-log 7`
D. `13.24+log 7`

Answer

To solve the problem step by step, we will follow the reasoning laid out in the video transcript and break it down into clear steps.

### Step 1: Write the Reaction
The reaction between TCl and NaOT can be written as:
\[ \text{TCl} + \text{NaOT} \rightarrow \text{NaCl} + \text{T}_2\text{O} \]

### Step 2: Calculate the Moles of TCl and NaOT
Using the formula for moles, which is:
\[ \text{Moles} = \text{Molarity} \times \text{Volume (in L)} \]

- For TCl:
\[ \text{Moles of TCl} = 0.2 \, \text{M} \times 0.010 \, \text{L} = 0.002 \, \text{mol} = 2 \, \text{mmol} \]

- For NaOT:
\[ \text{Moles of NaOT} = 0.25 \, \text{M} \times 0.015 \, \text{L} = 0.00375 \, \text{mol} = 3.75 \, \text{mmol} \]

### Step 3: Determine Remaining Moles After Reaction
After the reaction, the remaining moles of NaOT will be:
\[ \text{Remaining moles of NaOT} = 3.75 \, \text{mmol} - 2 \, \text{mmol} = 1.75 \, \text{mmol} \]

### Step 4: Calculate the Total Volume
The total volume of the solution after mixing is:
\[ \text{Total Volume} = 10 \, \text{mL} + 15 \, \text{mL} = 25 \, \text{mL} = 0.025 \, \text{L} \]

### Step 5: Calculate the Concentration of OH⁻ Ions
Using the formula for concentration:
\[ \text{Concentration of } \text{OH}^- = \frac{\text{Remaining moles of NaOT}}{\text{Total Volume}} = \frac{1.75 \, \text{mmol}}{25 \, \text{mL}} = \frac{1.75 \times 10^{-3} \, \text{mol}}{0.025 \, \text{L}} = 0.07 \, \text{M} \]

### Step 6: Calculate pOH
Using the formula for pOH:
\[ \text{pOH} = -\log[\text{OH}^-] = -\log(0.07) \approx 1.155 \]

### Step 7: Calculate pT
Using the relationship between pT, pOH, and the given pT of pure T₂O:
\[ \text{pT} + \text{pOH} = 14 \]
Given that the pT of pure T₂O is 7.62:
\[ \text{pT} + 1.155 = 14 \]
Thus:
\[ \text{pT} = 14 - 1.155 = 12.845 \]

### Final Answer
The pT of the solution prepared by adding 10 mL of 0.2 M TCl to 15 mL of 0.25 M NaOT is approximately **12.845**.

---