The enthalpy of neutraliztion of weak base A OH and a strong base BOH by HCl are -12250 cal/mol and -13000 cal/mol respectively . When one mole of HCl is added to a solution containting 1 mole of A OH and 1 mole of BOH , the enthalpy change was -12500 cal/mol . In what ratio is the acid distribution between A OH and BOH?

Options

A. `2 : 1`
B. `2 : 3`
C. `1 : 2`
D. None of these

Explanation

To solve the problem, we need to determine the ratio in which HCl is distributed between the weak base AOH and the strong base BOH when they are mixed together. We will use the given enthalpy changes of neutralization for both bases and the overall enthalpy change when HCl is added to the mixture.

### Step-by-Step Solution:

1. **Identify Variables:**
- Let \( x \) be the moles of HCl that react with AOH.
- Therefore, the moles of HCl that react with BOH will be \( 1 - x \) since we have 1 mole of HCl in total.

2. **Write the Enthalpy Change Equations:**
- The enthalpy change for the reaction of HCl with AOH is given as -12250 cal/mol.
- The enthalpy change for the reaction of HCl with BOH is given as -13000 cal/mol.
- The total enthalpy change when 1 mole of HCl is added to the mixture is -12500 cal/mol.

3. **Set Up the Equation:**
- The total heat released can be expressed as:
\[ \text{Total Heat} = (\text{Heat from AOH}) + (\text{Heat from BOH}) \]
- This can be written as:
\[ -12250x + (-13000)(1 - x) = -12500 \]

4. **Simplify the Equation:**
- Expanding the equation:
\[ -12250x - 13000 + 13000x = -12500 \]
- Combine like terms:
\[ 7500x - 13000 = -12500 \]
- Rearranging gives:
\[ 7500x = -12500 + 13000 \]
\[ 7500x = 500 \]

5. **Solve for \( x \):**
- Dividing both sides by 7500:
\[ x = \frac{500}{7500} = \frac{1}{15} \]

6. **Find \( 1 - x \):**
- Since \( 1 - x = 1 - \frac{1}{15} = \frac{14}{15} \).

7. **Determine the Ratio:**
- The ratio of the moles of HCl distributed to AOH and BOH is:
\[ \text{Ratio} = \frac{x}{1 - x} = \frac{\frac{1}{15}}{\frac{14}{15}} = \frac{1}{14} \]
- Thus, the ratio of AOH to BOH is \( 1:14 \).

### Final Answer:
The ratio of the acid distribution between AOH and BOH is \( 1:14 \).