`0.1` mol each of ethyl alcohol and acetic acid are allowed to react and at equilibrium, the acid was exactly neutralised by `100 mL` of `0.85 N NaOH`. If no hydrolysis of ester is supposed to have undergo, find `K_(c)`.

Explanation

To find the equilibrium constant \( K_c \) for the reaction between ethyl alcohol (ethanol) and acetic acid, we can follow these steps:

### Step 1: Write the Reaction
The reaction between acetic acid (CH₃COOH) and ethyl alcohol (C₂H₅OH) to form an ester (ethyl acetate, CH₃COOC₂H₅) and water (H₂O) can be represented as:

\[ \text{CH}_3\text{COOH} + \text{C}_2\text{H}_5\text{OH} \rightleftharpoons \text{CH}_3\text{COOC}_2\text{H}_5 + \text{H}_2\text{O} \]

### Step 2: Initial Moles
Initially, we have:
- Moles of acetic acid = 0.1 mol
- Moles of ethyl alcohol = 0.1 mol
- Moles of ester = 0
- Moles of water = 0

### Step 3: Change in Moles
Let \( x \) be the amount of acetic acid and ethyl alcohol that reacts at equilibrium. Therefore, at equilibrium, we have:
- Moles of acetic acid = \( 0.1 - x \)
- Moles of ethyl alcohol = \( 0.1 - x \)
- Moles of ester = \( x \)
- Moles of water = \( x \)

### Step 4: Neutralization with NaOH
We are given that 100 mL of 0.85 N NaOH was used to neutralize the acetic acid. The normality of NaOH indicates that 1 equivalent of NaOH reacts with 1 equivalent of acetic acid.

Calculating the milliequivalents of NaOH used:
\[ \text{Milliequivalents of NaOH} = \text{Volume (mL)} \times \text{Normality} = 100 \, \text{mL} \times 0.85 \, \text{N} = 85 \, \text{mEq} \]

Since acetic acid is monobasic, the milliequivalents of acetic acid that reacted is also 85 mEq.

### Step 5: Moles of Acetic Acid Left
The moles of acetic acid that reacted can be calculated as:
\[ \text{Moles of acetic acid that reacted} = \frac{85 \, \text{mEq}}{1000} = 0.085 \, \text{mol} \]

### Step 6: Calculate \( x \)
From the initial moles of acetic acid:
\[ 0.1 - x = 0.085 \]
Solving for \( x \):
\[ x = 0.1 - 0.085 = 0.015 \, \text{mol} \]

### Step 7: Equilibrium Concentrations
At equilibrium, we have:
- Moles of acetic acid = \( 0.1 - 0.015 = 0.085 \, \text{mol} \)
- Moles of ethyl alcohol = \( 0.1 - 0.015 = 0.085 \, \text{mol} \)
- Moles of ester = \( 0.015 \, \text{mol} \)
- Moles of water = \( 0.015 \, \text{mol} \)

### Step 8: Calculate \( K_c \)
The equilibrium constant \( K_c \) for the reaction is given by:
\[ K_c = \frac{[\text{Ester}][\text{Water}]}{[\text{Acetic Acid}][\text{Ethyl Alcohol}]} \]
Substituting the equilibrium concentrations:
\[ K_c = \frac{(0.015)(0.015)}{(0.085)(0.085)} \]
Calculating:
\[ K_c = \frac{0.000225}{0.007225} \approx 0.0311 \]

### Final Answer
Thus, the equilibrium constant \( K_c \) is approximately \( 0.0311 \).

---