0.1 mole of 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 undergone, find the equilibrium constant.

Answer

To solve the problem step-by-step, we will analyze the reaction between ethyl alcohol (C2H5OH) and acetic acid (CH3COOH) to form an ester (ethyl acetate, CH3COOC2H5) and water (H2O). The reaction can be represented as follows:

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

### Step 1: Determine Initial Moles
Initially, we have:
- Moles of ethyl alcohol = 0.1 moles
- Moles of acetic acid = 0.1 moles

### Step 2: Establish Equilibrium Conditions
At equilibrium, let \( x \) be the amount of acetic acid that has reacted. Therefore, the moles at equilibrium will be:
- Moles of ethyl alcohol = \( 0.1 - x \)
- Moles of acetic acid = \( 0.1 - x \)
- Moles of ester (ethyl acetate) = \( x \)
- Moles of water = \( x \)

### Step 3: Neutralization with NaOH
The problem states that the acetic acid was neutralized by 100 mL of 0.85 N NaOH. The normality (N) indicates the number of equivalents per liter.

Calculating the equivalents of NaOH used:
\[ \text{Equivalents of NaOH} = \text{Volume (L)} \times \text{Normality (N)} = 0.1 \, \text{L} \times 0.85 \, \text{N} = 0.085 \, \text{equivalents} \]

Since acetic acid is a weak acid and reacts in a 1:1 ratio with NaOH, the equivalents of acetic acid that remain unreacted at equilibrium is also 0.085 equivalents.

### Step 4: Relate Remaining Acetic Acid to Moles
The remaining moles of acetic acid can be calculated from the equivalents:
\[ \text{Remaining moles of acetic acid} = 0.085 \, \text{equivalents} = 0.085 \, \text{moles} \]

### Step 5: Set Up the Equation
From the initial moles of acetic acid (0.1 moles), we have:
\[ 0.1 - x = 0.085 \]
Solving for \( x \):
\[ x = 0.1 - 0.085 = 0.015 \, \text{moles} \]

### Step 6: Calculate Equilibrium Concentrations
Now we can find the equilibrium concentrations:
- Moles of ethyl alcohol at equilibrium = \( 0.1 - 0.015 = 0.085 \, \text{moles} \)
- Moles of acetic acid at equilibrium = \( 0.085 \, \text{moles} \)
- Moles of ester at equilibrium = \( 0.015 \, \text{moles} \)
- Moles of water at equilibrium = \( 0.015 \, \text{moles} \)

### Step 7: Calculate the Equilibrium Constant \( K_c \)
The equilibrium constant \( K_c \) for the reaction is given by:
\[ K_c = \frac{[\text{ester}][\text{water}]}{[\text{ethyl alcohol}][\text{acetic acid}]} \]
Substituting the equilibrium concentrations:
\[ K_c = \frac{(0.015)(0.015)}{(0.085)(0.085)} \]
Calculating \( K_c \):
\[ K_c = \frac{0.000225}{0.007225} \approx 0.0311 \]

### Final Answer
Thus, the equilibrium constant \( K_c \) is approximately \( 0.031 \).

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